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: \[ x = 0.025 { moles} \times 100 { g/mol} = 2.5 { g} \]