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SVEN [57.7K]
3 years ago
13

Part A

Chemistry
1 answer:
sertanlavr [38]3 years ago
5 0

Volume HNO₃ : 51.4 ml

Mass CO₂ : 0.17 g

<h3>Further explanation</h3>

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

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