A 49.3 sample of CaCO3 was treated with aqueous H2SO4, producing calcium sulfate, 3.65 g of water and CO2(g). What was the % yield of H2O?
a. 8.9
b. 58.9
c. 7.4
d. 92.6
e. 41.1
Guide On Rating System
Vote
To calculate the % yield of water (H2O), we need to compare the actual yield of water to the theoretical yield of water.
First, let's calculate the theoretical yield of water.
1 mol of CaCO3 produces 1 mol of H2O when it reacts with H2SO4.
The molar mass of CaCO3 is 100.09 g/mol.
The molar mass of H2O is 18.02 g/mol.
Therefore, 49.3 g of CaCO3 will produce (49.3 g / 100.09 g/mol) * (18.02 g/mol) = 8.918 g of water.
Now, let's calculate the actual yield of water, which is given as 3.65 g.
The % yield of water is calculated as (actual yield / theoretical yield) * 100%.
% yield of water = (3.65 g / 8.918 g) * 100% = 40.9%
Therefore, the % yield of water is 41.1%.
The correct answer is e. 41.1.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99