Answer:
CaCO3 is the limiting reactant
55 g of CO2 is made
Explanation:
First we must put down the reaction equation;
CaCO3(s) + 2HCl(aq) ---------> CaCl2(s) + H2O(l) + CO2(g)
Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles
From the reaction equation;
1 mole of CaCO3 yields 1 mole of CO2
Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2
For HCl;
number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles
From the reaction equation;
2 moles of HCl yields 1 mole of CO2
3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2
Hence CaCO3 is the limiting reactant.
Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2
I think the correct answer is D Hope this helps
