Question #2c774

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Question #2c774

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Answer 1 · 2017-12-05T12:08:03

Roughly $2491 k g$ $2491kg$

Explanation:

The reaction between $M g$ $Mg$ and $O_{2}$ $O_2$ to produce $M g O$ $MgO$ goes like this:

$2 M g_{(s)} + O_{2 (g)} \to 2 M g O_{(s)}$ $2Mg_((s))+O_(2(g))->2MgO_((s))$

$4131 k g = 4131000 g$ $4131kg=4131000g$

$n (M g O) = \frac{4131000}{24.3 + 16} = 102506.2035 m o l$ $n(MgO)=4131000/(24.3+16)=102506.2035mol$

Molar ratio of Mg:MgO is 1:1

So, $102506.2035 m o l$ $102506.2035mol$ of $M g$ $Mg$ are needed.

$M_{r} (M g) = 24.3$ $M_r(Mg)=24.3$

$m = n \cdot M_{r} = 24.3 (102506.2035) = 2490900.744 k g = 2490.900744 g \approx 2491 k g$ $m=n*M_r=24.3(102506.2035)=2490900.744kg=2490.900744g~~2491kg$

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Question #2c774

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