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Answer 1 · 2017-12-05T12:08:03

Roughly $2491kg$

The reaction between $Mg$ and $O_2$ to produce $MgO$ goes like this:

$2Mg_((s))+O_(2(g))->2MgO_((s))$

$4131kg=4131000g$

$n(MgO)=4131000/(24.3+16)=102506.2035mol$

Molar ratio of Mg:MgO is 1:1

So, $102506.2035mol$ of $Mg$ are needed.

$M_r(Mg)=24.3$

$m=n*M_r=24.3(102506.2035)=2490900.744kg=2490.900744g~~2491kg$