Question

stoichiometry-practice questions

1. when calcium carbonate reacts with hydrochloric acid, an aqueous solution of calcium chloride, water, and carbon dioxide gas is produced.

a. write and balance the chemical reaction associated with this reaction.
b. what is the mole ratio between the hydrochloric acid and the calcium chloride?
c. if 0.68 g hydrochloric acid reacts with excess calcium carbonate, what amount (in moles) of calcium chloride will be produced?

Explanation:

Part a:

Step1: Identify reactants and products

Reactants: Calcium carbonate ($\ce{CaCO3}$) and Hydrochloric acid ($\ce{HCl}$)
Products: Calcium chloride ($\ce{CaCl2}$), Water ($\ce{H2O}$), Carbon dioxide ($\ce{CO2}$)

Step2: Write unbalanced equation

$\ce{CaCO3 + HCl -> CaCl2 + H2O + CO2}$

Step3: Balance Cl atoms

There are 2 Cl in $\ce{CaCl2}$, so put 2 in front of $\ce{HCl}$:
$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$

Step4: Check other atoms (Ca, C, O, H)

• Ca: 1 on left, 1 on right (balanced)
• C: 1 on left, 1 on right (balanced)
• O: 3 in $\ce{CaCO3}$, 1 in $\ce{H2O}$ + 2 in $\ce{CO2}$ = 3 (balanced)
• H: 2 in $2\ce{HCl}$, 2 in $\ce{H2O}$ (balanced)

Step1: Identify coefficients from balanced equation

From $\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$, coefficient of $\ce{HCl}$ is 2, coefficient of $\ce{CaCl2}$ is 1.

Step2: Determine mole ratio

Mole ratio of $\ce{HCl}$ to $\ce{CaCl2}$ = $\frac{\text{Moles of } \ce{HCl}}{\text{Moles of } \ce{CaCl2}}$ = $\frac{2}{1}$

Step1: Calculate moles of HCl

Molar mass of $\ce{HCl}$ = $1.008 + 35.45 = 36.458$ g/mol
Moles of $\ce{HCl}$ = $\frac{\text{Mass of } \ce{HCl}}{\text{Molar mass of } \ce{HCl}}$ = $\frac{0.68\ \text{g}}{36.458\ \text{g/mol}} \approx 0.01865$ mol

Step2: Use mole ratio from balanced equation

From $\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$, mole ratio of $\ce{CaCl2}$ to $\ce{HCl}$ is $1:2$.

Step3: Calculate moles of $\ce{CaCl2}$

Moles of $\ce{CaCl2}$ = $\frac{1}{2} \times$ moles of $\ce{HCl}$
= $\frac{1}{2} \times 0.01865\ \text{mol} \approx 0.009325$ mol

Answer:

Balanced equation: $\boldsymbol{\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}}$

Part b: