Question

Answer the following for the reaction:
CaCO3(s)+2HCl(aq)→H2O(l)+CO2(g)+CaCl2(aq)

How many milliliters of a 0.240 M HCl solution can react with 9.25 g of CaCO3? Express your answer with the appropriate units.

Answer 1

Answer:

Explanation:

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio can be determined by examining the coefficients in front of formulas in a balanced chemical equation

Step 1: calculate the moles of CaCO₃

Given,

mass = 9.25 g

molar mass of  CaCO₃ = 100.0869 g/mol

we know,

moles = mass given / molar mass

moles of CaCO₃ = ( 9.25  g / 100.0869 g/mol ) = 0.09242 mol

Step 2: Calculate moles of HCl

CaCO₃(s) + 2 HCl(aq)-----> H₂O(l) + CO₂(g) + CaCl₂(aq)

According to the reaction:

1 mol of CaCO₃ completely reacts with 2 mol of HCl

So, 0.09242 mol of CaCO₃ will completely react with = (2 mol of HCl / 1 mol of CaCO₃ ) × 0.09242 mol of CaCO₃ = 0.18484 mol of HCl

Step 3: calculate the volume of HCl

we know

Molarity is defined to be the number of moles of solute divided by the volume of solution in liters

Molarity = moles / volume ( in L )

=> Volume ( in L ) = moles / molarity

we got , moles of HCl = 0.18484 mol

Molarity = 0.240 mol/L [ note: M = mol/L ]

on substituting the value

=> Volume ( in L ) = 0.18484 mol / 0.240 mol/L

=> Volume ( in L ) = 0.770164 L  ≈ 770.2 mL

Hence, the volume of HCl in milliliters required to react with 9.25 g = 770.2 mL

[ note :   1 L = 1000 mL hence, 0.770164 L =( 0.770164 L  × (1000 mL / 1 L )) ≈ 770.2 mL ]