How many grams of lead (II) chloride (PbCl2) are produced from the reaction of 7.000 g of NaCl and 50.400 g of Pb(NO3)2?Calculate the theoritical yield of PbCl2? (Atomic mass: Na-22.990 g/mol; 0-15.999 g/mol; Cl-35.453g/mol; Pb-207.200 g/mol; N-14.007).
2NaCl+ Pb(NO3)2 → 2 NaNO3 + PbCl2
21.136 g
16.547 g
33.373 g
42.272 g
To solve this problem, we need to use stoichiometry to determine the limiting reagent and the theoretical yield of PbCl2.
First, we need to find the number of moles of each reactant:
- Moles of NaCl = 7.000 g / 22.990 g/mol = 0.304 mol
- Moles of Pb(NO3)2 = 50.400 g / 331.200 g/mol = 0.152 mol
Next, we need to use the balanced equation to determine the mole ratio between NaCl and PbCl2:
- 2 moles of NaCl produce 1 mole of PbCl2
Now we can determine which reactant is limiting:
- If all the NaCl reacted, it would produce: 0.304 mol NaCl x (1 mol PbCl2 / 2 mol NaCl) = 0.152 mol PbCl2
- If all the Pb(NO3)2 reacted, it would produce: 0.152 mol Pb(NO3)2 x (1 mol PbCl2 / 1 mol Pb(NO3)2) = 0.152 mol PbCl2
Since the moles of PbCl2 produced are the same in both calculations, this means that NaCl is the limiting reagent.
Finally, we can use the mole ratio between NaCl and PbCl2 to calculate the theoretical yield of PbCl2:
- Theoretical yield of PbCl2 = 0.304 mol NaCl x (1 mol PbCl2 / 2 mol NaCl) x 239.200 g/mol PbCl2 = 21.136 g PbCl2
Therefore, the correct answer is (A) 21.136 g.
23,780 g of acetaldehyde (CH3CHO) reacts completely with oxygen to produce acetic acid (HC2H3O2) using manganese (II) acaetae catalyst according to the equation shown : 2CH3CHO + O2 = 2HC2H3O2
What mass of oxygen reacted with acetaldehyde?
First, we need to use stoichiometry to determine the moles of oxygen that react with acetaldehyde:
- From the balanced equation, we know that 2 moles of acetaldehyde react with 1 mole of oxygen.
- We can calculate the moles of acetaldehyde present by dividing its mass by its molar mass: 23.780 g / 44.053 g/mol = 0.539 mol.
- Using the mole ratio from the balanced equation, we can calculate the moles of oxygen needed: 0.539 mol acetaldehyde x (1 mol oxygen / 2 mol acetaldehyde) = 0.270 mol oxygen.
Finally, we can convert the moles of oxygen to its mass using its molar mass:
- Mass of oxygen = 0.270 mol x 32.00 g/mol = 8.64 g.
Therefore, the mass of oxygen that reacted with acetaldehyde is 8.64 g.
Give the answer to the correct significant digit 0.+71+96.6+2.003 + 0.0054 + 2.1 =
102.78
A student performed an analysis of a sample for its calcium content and got the following results: 14.91% 14.89%, 14.88%, and 14.90%. The actual amount of calcium in the sample is 14.90%. Calculate the percentage error and precision.
Comment on the precision and accuracy of the results.
To calculate the percentage error:
- Subtract the actual value from each of the measured values: 14.91% - 14.90% = 0.01%, 14.89% - 14.90% = -0.01%, 14.88% - 14.90% = -0.02%, and 14.90% - 14.90% = 0%.
- Take the absolute value of each of these differences: 0.01%, 0.01%, 0.02%, and 0%.
- Calculate the average of these absolute differences: (0.01% + 0.01% + 0.02% + 0%) / 4 = 0.01%
- The percentage error is the absolute difference between the actual value and the average of the measured values, divided by the actual value, multiplied by 100%: |14.90% - 14.895%| / 14.90% x 100% = 0.033% (rounded to 3 significant figures)
To calculate the precision:
- Calculate the range of the measured values: 14.91% - 14.88% = 0.03%
- The precision can be expressed as the range divided by the average of the measured values, multiplied by 100%: 0.03% / ((14.91% + 14.89% + 14.88% + 14.90%) / 4) x 100% = 0.20% (rounded to 2 significant figures)
Based on these calculations, we can see that the percentage error is quite small (0.033%), which suggests that the accuracy of the results is good. However, the precision is relatively poor (0.20%), which suggests that the measurements are not very consistent. Due to the small size of the sample, it is possible that the precision could be improved with more replicates (i.e. more measurements). Overall, the results are fairly accurate but not very precise.
James mistakenly mixed hexane with dichloromethane and water. If the density of hexane is 661 kg/m³, the density of dichloromethane is 1322 kg/m³, and the density of water is 1000 kg/m³, what is the best technique for separating the mixtures
The best technique for separating the mixture depends on the properties of the components and the intended application of the separated substances. However, the densities of the three substances suggest that a separation technique based on density differences could be effective.
Based on the densities provided, one possible separation technique that could work is liquid-liquid extraction.
To separate the hexane, dichloromethane, and water mixture using liquid-liquid extraction, the following steps could be taken:
1. Add a polar solvent, such as methanol or ethanol, to the mixture. The polar solvent will selectively dissolve the water, but not the hexane or dichloromethane.
2. After mixing thoroughly, allow the mixture to settle. The polar solvent and water will form a separate layer on top of the hexane and dichloromethane layer, due to their differing densities.
3. Carefully separate the two layers by decantation or using a separatory funnel. The hexane and dichloromethane can be collected in one container, while the polar solvent and water can be collected in another container.
4. The polar solvent and water can be further separated by distillation, since they have different boiling points.
Alternatively, if it is desired to separate the hexane and dichloromethane from the water, rather than the other way around, a technique such as fractional distillation could be used. Fractional distillation takes advantage of the different boiling points of the components to separate them based on their vapor pressures.
In either case, the densities of the substances can provide some guidance when selecting an appropriate separation technique, but other factors such as polarity, boiling points, and solubility should also be considered.
A metal carbonate, XCO3 of mass 2.012g was heated resulting in formation of XO, a metal oxide and carbon dioxide with a mass of 0.855g according to the reaction shown : 1.XCO3(s) = XO (s) + CO2 (g)
Atomic mass of O-15.999g/mol; H-1.008g/mol;C-12.011g/mol). Hint; calculate the mole of CO2 first.
To solve this problem, we need to use stoichiometry and the given masses to find the molar mass of the metal carbonate, and then use that information to determine the identity of the metal, X.
First, we need to calculate the number of moles of carbon dioxide produced:
- The mass of carbon dioxide produced is 0.855 g.
- Using the molar mass of carbon dioxide (44.01 g/mol), we can calculate the number of moles: 0.855 g / 44.01 g/mol = 0.0194 mol.
Next, we need to use the balanced equation to determine the number of moles of XCO3 that must have reacted to produce this amount of carbon dioxide:
- The balanced equation tells us that 1 mole of XCO3 produces 1 mole of CO2.
- Therefore, the number of moles of XCO3 that reacted is also 0.0194 mol.
Now we can use the mass of the metal carbonate and the number of moles that reacted to calculate the molar mass of XCO3:
- The mass of XCO3 is 2.012 g.
- The number of moles of XCO3 is 0.0194 mol.
- Therefore, the molar mass of XCO3 is: 2.012 g / 0.0194 mol = 103.6 g/mol.
The molar mass of XCO3 is a clue to the identity of the metal, X. Looking at the periodic table, we can see that there are a few metals with molar masses close to 103.6 g/mol, including manganese (54.94 g/mol) and iron (55.85 g/mol). However, these metals typically form oxides that are not solid at room temperature. The oxide formed in this reaction, XO, is a solid, which suggests that X is one of the alkali or alkaline earth metals, which commonly form solid oxides.
The molar mass of XCO3 corresponds to a metal with an atomic mass close to 40 g/mol. This could be either calcium (40.08 g/mol) or strontium (87.62 g/mol). However, since calcium is more abundant and more commonly found in nature, it is likely that X is calcium.
Therefore, the metal carbonate is likely calcium carbonate, CaCO3. The reaction can be written as: CaCO3(s) → CaO(s) + CO2(g).
Note: It is important to check the results of our calculations to ensure that the mass balance is correct. In this case, the mass of the products (0.855 g + mass of XO) must equal the mass of the reactant (2.012 g). We can check this by subtracting the mass of carbon dioxide from the total mass of the products: 2.012 g - 0.855 g = 1.157 g. This is the mass of the metal oxide, XO.
An unknown compound contains 13.04% of hydrogen, 34,78% of oxygen, and 52.17% of carbon by mass. Determine the empirical formula of the unknown compound
To determine the empirical formula of the unknown compound, we need to convert the percentages of each element into masses, then calculate the number of moles of each element, and finally divide each number of moles by the smallest number of moles to obtain whole number ratios. These ratios correspond to the subscripts in the empirical formula.
Assuming a 100 g sample, we can calculate the masses of each element:
- Mass of hydrogen (H) = 13.04 g
- Mass of oxygen (O) = 34.78 g
- Mass of carbon (C) = 52.17 g
We can then calculate the number of moles of each element using their atomic masses (H: 1.008 g/mol, C: 12.011 g/mol, O: 15.999 g/mol):
- Moles of hydrogen = 13.04 g / 1.008 g/mol = 12.93 mol
- Moles of oxygen = 34.78 g / 15.999 g/mol = 2.17 mol
- Moles of carbon = 52.17 g / 12.011 g/mol = 4.34 mol
Next, we need to find the simplest whole number ratio by dividing each number of moles by the smallest number of moles:
- Dividing all numbers of moles by 2.17, the smallest number of moles, gives the following:
- Hydrogen: 12.93 mol / 2.17 mol = 5.96
- Oxygen: 2.17 mol / 2.17 mol = 1.00
- Carbon: 4.34 mol / 2.17 mol = 2.00
Rounding to the nearest whole number, we get the empirical formula as H6C2O.
Therefore, the empirical formula of the unknown compound is H6C2O.
The pressure of a sample of an inert gas reduced significantly from 1000 kPa to 30kPa. The temperature increases from 27 degrees Celsius to 93 degrees Celsius. If the initial volume is 20 L, what is the final volume?
To solve this problem we can use the combined gas law, which relates the pressure, volume, and temperature of a gas. The formula is:
P1V1/T1 = P2V2/T2
where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.
We can rearrange this formula to solve for the final volume, V2, as follows:
V2 = (P1V1T2) / (T1P2)
Substituting the given values into this equation, we get:
V2 = (1000 kPa)(20 L)(366 K) / (300 K)(30 kPa)
Simplifying, we get:
V2 = (1000/30)(20/1)(366/300) L
V2 = 244 L
Therefore, the final volume is approximately 244 L.