1. What weight of aluminum may be obtain from 1 kg of aluminum oxide?
2. how many liters of hydrogen would be form by the reaction of 25g of hydrochloric acid and sufficient quantity of zinc?
3. 0.82l of an unknown gas at 35 degree Celsius and 80 cm-Hg weigh 2.46g. compute for the molecular weight of the unknown gas?
1.) The chemical formula of aluminum oxide is Al2O3. The total mass per mole is
27 + 3*16 = 75 g. Of that mass, 27 g is aluminum atoms. Therefore 27/75 or 36.0% of the mass is aluminum.
The answer is therefore 36% of 1000 g. How much is that?
2.) First, let's write down the chemical reaction.
Zn + 2 HCl -> ZnCl2 + H2
25 g of HCl is 25/36.5 = 0.685 moles.
that will form, wityh excess Zn, half as many moles of H2. Convert that to gas volume assuming STP conditions.
3. 80 cm Hg is the PRESSURE, and that equals = 0.1053 atm. At 35 deg C (which is T = 308 K) the number of moles in 0.82 L would be given by
n = PV/RT = (0.1053*0.82)/(0.08206*308) = 3.42*10^-3 moles
If that is 26 g, the mass of one mole is 7600 g
That is a very large molecular mass. Are you sure all of your numbers are correct?
1. To determine the weight of aluminum obtained from 1 kg of aluminum oxide, we need to consider the molar mass and stoichiometry of the reaction involved.
Firstly, we must determine the molar mass of aluminum oxide (Al2O3). The atomic mass of aluminum (Al) is approximately 26.98 g/mol, and the atomic mass of oxygen (O) is approximately 16.00 g/mol. Since there are two aluminum atoms and three oxygen atoms in one molecule of aluminum oxide, the molar mass can be calculated as follows:
(2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol
So, one mole of aluminum oxide weighs approximately 101.96 g.
Next, we need to consider the stoichiometry of the reaction involved. From the balanced chemical equation:
2 Al2O3 => 4 Al + 3 O2
It shows that for every two moles of aluminum oxide, we obtain four moles of aluminum.
Therefore, using the concept of molar mass, we can determine the amount of aluminum formed from 1 kg of aluminum oxide:
(4 moles Al / 2 moles Al2O3) * (101.96 g/mole) = 203.92 g
Hence, approximately 203.92 grams of aluminum can be obtained from 1 kg of aluminum oxide.
2. To determine the volume of hydrogen gas produced from the reaction between hydrochloric acid (HCl) and zinc (Zn), we need to consider the balanced chemical equation and stoichiometry.
The balanced chemical equation for this reaction is:
Zn + 2 HCl => ZnCl2 + H2
From the equation, we can see that one mole of zinc reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas.
Firstly, we need to determine the molar mass of hydrochloric acid (HCl). The atomic mass of hydrogen (H) is approximately 1.01 g/mol, and the atomic mass of chlorine (Cl) is approximately 35.45 g/mol. Hence, the molar mass of HCl is 1.01 g/mol + 35.45 g/mol = 36.46 g/mol.
Now, calculate the number of moles of HCl present in 25 g:
Number of moles = mass / molar mass = 25 g / 36.46 g/mol ≈ 0.686 moles
Since the stoichiometry of the reaction tells us that 1 mole of zinc reacts with 2 moles of HCl, we can determine the number of moles of hydrogen gas produced as follows:
0.686 moles HCl * (1 mole H2 / 2 moles HCl) = 0.343 moles H2
Finally, to calculate the volume of hydrogen gas at standard temperature and pressure (STP), you can use the ideal gas law. At STP, 1 mole of any gas occupies a volume of approximately 22.4 liters.
Hence, the volume of hydrogen gas produced would be:
0.343 moles H2 * 22.4 L/mole ≈ 7.68 liters
Therefore, approximately 7.68 liters of hydrogen gas would be formed from the reaction.
3. To find the molecular weight of the unknown gas, we can use the ideal gas law equation:
PV = nRT
where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant (0.0821 L.atm/K.mol), and T represents temperature in Kelvin.
First, convert the temperature from Celsius to Kelvin:
35°C + 273.15 = 308.15 K
Next, convert the pressure from cm-Hg to atm:
80 cm-Hg * (1 atm / 76 cm-Hg) = 1.053 atm
Using the ideal gas law equation, we can calculate the number of moles of the unknown gas:
n = PV / RT = (1.053 atm * 0.82 L) / (0.0821 L.atm/K.mol * 308.15 K) ≈ 0.034 moles
Finally, to calculate the molecular weight (molar mass) of the unknown gas, divide the mass by the number of moles:
molecular weight = mass / moles = 2.46 g / 0.034 moles ≈ 72.35 g/mol
Therefore, the molecular weight of the unknown gas is approximately 72.35 g/mol.