Question

You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 74.24 g . The relevant equation is CaCO_3(s) + 2HCl(aq) rightarrow H_2O(l) + CO_2(g) + CaCl_2(aq) Assuming no other reactions take place, what mass of CO_2 was produced in this reaction?

Answer 1

Answer: Mass of $CO_2$  produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)$

Mass or reactants =  Mass of $CaCO_3$+ mass of $HCl$ = 16.00 + 64.80 = 80.80 g

Mass of products  = mass of aqueous solution + mass of $CO_2$ + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of $CO_2$  produced in this reaction was 6.56 grams