2. A 0.6911-g sample of impure strontium carbonate, SrCO3, isdecomposed with HCl

Question

2. A 0.6911-g sample of impure strontium carbonate, SrCO3, isdecomposed with HCl. The resulting liberated CO2 is
collected on calcium oxide and weighs 0.1745 g. Calculate thepercentage of SrCO3 in the sample. Hint: write the
reaction equation, thinking about what would be the most likelystrontium product, considering the reactants...

Explanation / Answer

First balance theequation SrCO3 (aq) + 2HCl (aq) ---> SrCl (aq) + CO2 +H2O (Carbonic acid is formed in the reaction which readily decomposesto carbon dioxide and water) Given 1:2:1:1:1 stoichiometric ratio 0.1745 g CO2 produced 0.6911 g impure SrCO3 sample MMCO2 = 44.010 g/mol MMSrCO3 = 147.6289 g/mol Calculate the amount (in grams) ofSrCO3 that would theoretically yield 0.1745 gCO2 (0.1745 g CO2)(1 mol CO2 / 44.010g CO2)(1 mol SrCO3 / 1mol CO2)(147.6289 g SrCO3 / 1 molSrCO3) = 0.5853 g SrCO3 Now calculate the percent by mass ofSrCO3 in the sample % mass = (mass of component / mass of sample) x 100 = (0.5853g / 0.6911 g) x 100 =84.69 %SrCO3

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