Question #7607c

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Question #7607c

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Answer 1 · 2015-03-17T00:41:47

We do all this by ratios $2 H_{2} + O_{2} \to 2 H_{2} O$ $2H_2+O_2->2H_2O$

Molecular mass:
$H_{2} \to 2 \cdot 1 = 2$ $H_2->2*1=2$ times 2= $4$ $4$
$O_{2} \to 2 \cdot 16 = 32$ $O_2->2*16=32$
$H_{2} O \to 2 \cdot 1 + 1 \cdot 16 = 18$ $H_2O->2*1+1*16=18$ times 2= $36$ $36$
So the weight ratio of this reaction is
$4 + 32 \to 36 or 1 + 8 \to 9$ $4+32->36or1+8->9$

Now we calculate how much $O_{2}$ $O_2$ is needed for the $H_{2}$ $H_2$ :
$\frac{16.8}{1} = \frac{x}{8} \to x = 134.4 g$ $16.8/1=x/8->x=134.4g$ which we don't have.

The other way around:
$\frac{x}{1} = \frac{46.4}{8} \to x = 5.8 g$ $x/1=46.4/8->x=5.8g$

So $5.8 g$ $5.8g$ of $H_{2}$ $H_2$ will react with all $46.4 g$ $46.4g$ of $O_{2}$ $O_2$

This should give $5.8 + 46.4 = 52.2 g$ $5.8+46.4=52.2g$ of $H_{2} O$ $H_2O$

The percentage yield is thus:
$\frac{12}{52.2} \cdot 100 \approx 23 %$ $12/52.2 *100~~23%$

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Question #7607c

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