# Can you check my working for this stoichiometry problem?

Anonymous
timer Asked: Dec 29th, 2018

Question description

Q: One impure sample of CaCO3 is treated with 100.00mL of 1.00mol/L HCl. The resulting solution is diluted to 250.00 ML and 25.00 ML of this solution required 20.00mL of 0.10mol/L NaOH for neutralisation. What is the mass of CaCO3 assuming impurities are not reactive substances?

Working:

Balanced equation:

CaCO3 + 2HCl > CaCl2 + CO2 + H2O

1.00 mol x .100 L = 0.1 M / .250 L = 0.4 M

(1/10) x .250 L = .025 L

0.4 M / 10 = 0.04 M

0.04 M x .5 = 0.02 M

0.02 M x .025 L = 0.001 mol

0.001 mol x 100.0896 g (CaCO3 molar mass) = 0.1 g

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