Question

50 mL of 1 M HCl was completely reacted with $x$ g of CaCO3 to form CaCl2, CO2, and H2O. What is the value of $x$ in g? 
 

• A. 25
• B. 0.25
• C. 2.5
• D. 0.025

Hint:

Remember to balance chemical equations correctly and relate the stoichiometry to the amount of reactants used or products formed in a reaction.

Answer 1

The Correct Option is C

First, we calculate the number of moles of HCl: $${Moles of HCl} = 1 { M} \times 0.05 { L} = 0.05 { moles}$$ The reaction between HCl and CaCO3 is as follows: $${CaCO}_3 + 2 {HCl} \rightarrow {CaCl}_2 + {CO}_2 + {H}_2{O}$$ For every mole of CaCO3, 2 moles of HCl are required. Thus, the moles of CaCO3 that reacted: $${Moles of CaCO}_3 = \frac{0.05 { moles HCl}}{2} = 0.025 { moles}$$ The molar mass of CaCO3 is approximately 100 g/mol, so: $$x = 0.025 { moles} \times 100 { g/mol} = 2.5 { g}$$