Explanation:
(a) write a balanced equation for the reaction
CaCO3 + HCl --> CaCl2 + H2O + CO2
The balanced equation is given as;
CaCO3 + 2HCl → CaCl2 + H2O + CO2
(b) when the reaction was complete, 800 mL of carbon dioxide gas was collected. How many moles of calcium carbonate were used in the creation?
From the balanced reaction;
1 mol of CaCO3 reacts to produce 1 mol of CO2
1 mol of CO2 = 22.4 L of CO2
This means;
1 mol of CaCO3 reacts to produce 22.4L of CO2
x mol would produce 800ml (0.8 L) of CO2
1 = 22.4
x = 0.8
x = 0.8 * 1 / 22.4 = 0.0357 mol
(c) How many grams of CaCO3 were used?
Mass = Number of moles * Molar mass
Molar mass of CaCO3 = 100.0869 g/mol
Mass = 0.0357 mol * 100.0869 g/mol = 3.57 g
(d) If there was another contaminant in the sample that was not un reactive, would this have caused the percent yield of carbon dioxide to be higher, lower, or the same, explain your answer.
The same
An un reactive contaminant in the sample is most likely a catalyst. Catalysts only affect the rate of reaction. They do not affect yields of products.
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