Question #b58ba

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Question #b58ba

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Answer 1 · 2017-12-22T13:29:46

A) 22g

Explanation:

Calculate this using the equation:

$M o l = \frac{M a s s}{M r}$ $Mol = (Mass) / (Mr)$

Mr $C O$ $CO$ = 28 g/mol
Mr $C O_{2}$ $CO_2$ = 44 g/mol

$\frac{14}{28} = 0.5$ $14 / 28 = 0.5$ mol

Based on the balanced equation you provided, $C O$ $CO$ and $C O_{2}$ $CO_2$ are in a 1:1 ratio. Therefore, the number of moles of $C O$ $CO$ used = the number of moles of $C O_{2}$ $CO_2$ produced.

Rearranging the equation for mass

$M a s s = M o l x M r$ $Mass = Mol x Mr$

$0.5 \times 44 = 22 g$ $0.5 xx 44 = 22g$

Answer 2 · 2017-12-22T13:30:21

The answer is option $(A) = 22 g$ $(A)=22g$

Explanation:

The reaction is

$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g)+O_2(g)rarr 2CO_2(g)$

$2$ $2$ moles of carbon monoxide give $2$ $2$ moles of carbon dioxide

$1$ $1$ mole of carbon monoxide is $= 12 + 16 = 28 g$ $=12+16=28g$

$1$ $1$ mole of carbon dioxide is $= 12 + 2 \cdot 16 = 44 g$ $=12+ 2*16=44g$

Therefore,

The mass of carbon dioxide is $= \frac{44}{28} \cdot 14 = 22 g$ $=44/28*14=22g$

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Question #b58ba

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