A 211 g sample of barium carbonate reacts with a solution of nitric acid to give barium nitrate, carbon dioxide, and water. If the acid is present in excess, what mass and volume of carbon dioxide gas at STP will form?

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A 211 g sample of barium carbonate reacts with a solution of nitric acid to give barium nitrate, carbon dioxide, and water. If the acid is present in excess, what mass and volume of carbon dioxide gas at STP will form?

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Answer 1 · 2017-05-18T06:03:32

We assess the stoichiometric equation...............

Explanation:

$B a C O_{3} (s) + 2 H N O_{3} (a q) \to B a {(N O_{3})}_{2} (a q) + H_{2} O (l) + C O_{2} (g) ↑ ⏐$ $BaCO_3(s) + 2HNO_3(aq) rarrBa(NO_3)_2(aq)+ H_2O(l) + CO_2(g)uarr$

Given the stoichiometry, ONE MOLE of carbon dioxide is evolved for each mole of barium carbonate.

$Moles of barium carbonate = \frac{211 \cdot g}{197.34 \cdot g \cdot m o l^{- 1}} = 1.07 \cdot m o l .$ $"Moles of barium carbonate"=(211*g)/(197.34*g*mol^-1)=1.07*mol.$

Given the stoichiometry, $1.07 \cdot m o l$ $1.07*mol$ of carbon dioxide gas will be evolved.

With respect to $C O_{2}$ $CO_2$ , this represents a mass of $1.07 \cdot m o l \times 44.01 \cdot g \cdot m o l^{- 1} = ? ? \cdot g .$ $1.07*molxx44.01*g*mol^-1=??*g.$

And a volume of $1.07 \cdot m o l \times 24.5 \cdot L \cdot m o l^{- 1}$ $1.07*molxx24.5*L*mol^-1$ given room temperature and pressure...

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A 211 g sample of barium carbonate reacts with a solution of nitric acid to give barium nitrate, carbon dioxide, and water. If the acid is present in excess, what mass and volume of carbon dioxide gas at STP will form?

1 Answer

Explanation:

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