4.How many grams of calcium chloride can be prepared if a student uses 500g if calcium carbonate and 2.00L of a 4.00 Mol/L sodium chloride solution in the reaction.??
5. 2KMnoO4 + 5 H2C2O4<----need to use this
Concentration of 0.24L KMnO4 solution required to oxidize 2.7g of Oxalic Acid???
6.What volume of 0.100mol/L aqueous sodium hydroxide(NaOH) is neutralized by 500ml of 0.200mol/L hydrochloric acid(HCI) ??
7.When 10.0g of potassium chlorate,KCIO3,decomposes,what volume of oxygen is obtained(at 18C and 750 mm HG pressure) ??
4.How many grams of calcium chloride can be prepared if a student uses 500g if calcium carbonate and 2.00L of a 4.00 Mol/L sodium chloride solution in the reaction.??
Alex, could you check the question?
Calcium carbonate does not react directly with sodium chloride to make calcium chloride, just like sea water does not dissolve lime-stone. Otherwise many parts of the world would be eroded by the sea. However, it is easily possible to make calcium chloride from hydrochloric acid through the reaction:
CaCO3 + 2HCl -> CaCl2 + H2O
5. 2KMnoO4 + 5 H2C2O4<----need to use this
Concentration of 0.24L KMnO4 solution required to oxidize 2.7g of Oxalic Acid???
This is a complex redox reaction in which the potassium permanganate oxidizes the oxalic acid in an acidic solution (presence of H+ ions.
The net reaction is shown below, where the H+ ions would normally supplied by sulphuric acid or another strong acid.
5H2C2O4 (aq) + 6H+ + 2MnO4- (aq) --> 10CO2 (g) + 2 Mn2+(aq) +8H20 (l)
This is a very interesting titration where there is an effect of autocatalysis and no indicator is required because of the colour of potassium permanganate. Read up the following articles for your interest and information:
http://jchemed.chem.wisc.edu/JCeSoft/CCA/CCA3/MAIN/AUTOCAT/PAGE1.HTM
http://www.peerpapers.com/essays/Determination-Molarity-Permanganate-Solution-Via-Titration/40919.html?read_essay
By using the ratio of molecular weights of 2KMnoO4 + 5 H2C2O4, you should have no problem calculating the molarity of KMNO4 required. Refer to solution for questions 1-3 for how this can be done.
(ref:http://www.jiskha.com/display.cgi?id=1244169404)
6.What volume of 0.100mol/L aqueous sodium hydroxide(NaOH) is neutralized by 500ml of 0.200mol/L hydrochloric acid(HCI) ??
NaOH + HCl --> NaCl + H2O
From
500 ml of 0.2 mol/l of HCl contains
(500/1000)*0.2=0.1 mole of HCl,
and
1 litre of 0.1 mol/l of NaOH contains 0.1 mole of NaOH,
we conclude that 1 litre of 0.1 mol/l NaOH aqueous solution is required.
7.When 10.0g of potassium chlorate,KCIO3,decomposes,what volume of oxygen is obtained(at 18C and 750 mm HG pressure) ??
2KClO3 + heat -> 2KCl + 3O2
Managanese dioxide is used as a catalyst for this reaction.
Molecular weight of KClO3
=39+35.5+3*16=122.5
Number of moles of oxygen produced
=10/(2*122.5)*3
=0.1224 mole
Volume of oxygen
= 0.1224*22.4 l.
= 2.74 litres
Erratum: Manganese dioxide
yes..for #4 it is hydrochloric acid i forgot to put it in,how would u find the grams of calcium chloride???Thank you MathMate
4. To find the number of grams of calcium chloride that can be prepared, you will need to calculate the amount of calcium chloride produced in the reaction between calcium carbonate and sodium chloride.
First, determine the molar mass of calcium carbonate (CaCO3) and sodium chloride (NaCl) using the periodic table. The molar mass of CaCO3 is approximately 100.1 g/mol and NaCl is approximately 58.4 g/mol.
Next, use stoichiometry to find the molar ratio between calcium carbonate and calcium chloride in the balanced equation. The balanced equation for the reaction is:
CaCO3 + 2NaCl -> CaCl2 + Na2CO3
From the equation, you can see that one mole of calcium carbonate produces one mole of calcium chloride. This means that the number of moles of calcium chloride produced will be the same as the number of moles of calcium carbonate.
To find the number of moles of calcium carbonate, divide the mass of calcium carbonate (500 g) by its molar mass (100.1 g/mol). This will give you the number of moles of calcium carbonate.
Next, calculate the volume of sodium chloride using the given concentration (4.00 mol/L) and the given volume (2.00 L). The concentration is the number of moles of sodium chloride per liter of solution, so multiplying the concentration by the volume will give you the number of moles of sodium chloride.
Now, you have the number of moles of calcium carbonate and the number of moles of sodium chloride. Based on the balanced equation, these numbers of moles are equal to the number of moles of calcium chloride produced.
Finally, to find the mass of calcium chloride, multiply the number of moles of calcium chloride by its molar mass (approximately 110.98 g/mol). This will give you the mass of calcium chloride that can be prepared using 500 g of calcium carbonate and 2.00 L of a 4.00 mol/L sodium chloride solution.
5. To determine the concentration of the KMnO4 solution needed to oxidize a given amount of oxalic acid (H2C2O4), you need to use the balanced equation of the reaction and the stoichiometry.
The balanced equation for the reaction is:
2KMnO4 + 5H2C2O4 -> 2MnC2O4 + K2C2O4 + 3H2O + 2CO2
From the equation, you can see that the ratio between KMnO4 and H2C2O4 is 2:5. This means that for every 2 moles of KMnO4, you need 5 moles of H2C2O4.
First, calculate the number of moles of oxalic acid using the given mass (2.7 g) and its molar mass (approximately 90.03 g/mol).
Next, apply the stoichiometry ratio to find the number of moles of KMnO4 needed. Divide the number of moles of H2C2O4 by 5 and multiply the result by 2.
Finally, to find the volume of the KMnO4 solution, divide the number of moles of KMnO4 needed by the concentration of the solution. The concentration is given as 0.24 L, which is its volume, so you can use this value directly.
6. To find the volume of 0.100 mol/L NaOH solution neutralized by 500 mL of 0.200 mol/L HCl solution, you need to use the balanced equation and apply stoichiometry.
The balanced equation for the neutralization reaction between NaOH and HCl is:
NaOH + HCl -> NaCl + H2O
From the equation, you can see that the ratio between NaOH and HCl is 1:1. This means that for every mole of NaOH, you need one mole of HCl.
First, calculate the number of moles of HCl using the given volume (500 mL) and its concentration (0.200 mol/L). Convert the volume to liters by dividing it by 1000, then multiply the result by the concentration.
Next, apply the stoichiometry ratio to find the number of moles of NaOH needed. Since the ratio is 1:1, the number of moles of NaOH needed will be equal to the number of moles of HCl.
Finally, to find the volume of the NaOH solution, divide the number of moles of NaOH needed by its concentration (0.100 mol/L). The result will be the volume of the NaOH solution required to neutralize the given volume of HCl solution.
7. To calculate the volume of oxygen obtained when 10.0 g of potassium chlorate (KClO3) decomposes, you need to know the balanced equation for the decomposition reaction and the ideal gas law.
The balanced equation for the decomposition of KClO3 is:
2KClO3 -> 2KCl + 3O2
From the equation, you can see that the ratio between KClO3 and O2 is 2:3. This means that for every 2 moles of KClO3, you obtain 3 moles of O2.
First, calculate the number of moles of KClO3 using the given mass (10.0 g) and its molar mass (approximately 122.55 g/mol).
Next, apply the stoichiometry ratio to find the number of moles of O2 obtained. Divide the number of moles of KClO3 by 2 and multiply the result by 3.
Now, you can use the ideal gas law to find the volume of O2 gas. The ideal gas law equation is:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given that the pressure is 750 mmHg (which you can convert to atm), you can rearrange the equation to solve for V:
V = (nRT) / P
Substitute the values you have so far into the equation (n, R, P) and convert the temperature from Celsius to Kelvin (18°C + 273 = 291 K). This will give you the volume of O2 gas obtained at the given pressure and temperature.