Question

Part A

What volume (in mL ) of a 0.150 M HNO3 solution will completely react with 35.7 mL of a 0.108 M Na2CO3 solution according to the following balanced chemical equation?

In the reaction in Part A, what mass (in grams) of carbon dioxide forms?

Answer 1

Volume HNO₃ : 51.4 ml

Mass CO₂ : 0.17 g

Further explanation

Reaction

Na₂CO₃ + 2 HNO₃ ⇒ 2 NaNO₃ + CO₂ + H₂O

V Na₂CO₃ = 35.7 ml

M  Na₂CO₃ = 0.108

mol Na₂CO₃ :

$\tt 35.7\times 0.108=3.8556~mlmol$

mol ratio Na₂CO₃  : HNO₃ = 1 : 2

mol HNO₃ :

$\tt \dfrac{2}{1}\times 3.8556=7.71~mlmol$

Volume HNO₃ :

$\tt mol=n=M\times V\\\\V=\dfrac{n}{M}=\dfrac{7.71}{0.15}=51.4~ml$

mol CO₂ = mol Na₂CO₃ = 3.8556 mlmol=0.00386 mol

mass CO₂(MW=44 g/mol) :

$\tt 0.00386\times 44=0.17~g$