Six grams each of magnesium and oxygen were allowed to react. Assuming that the reaction is complete, the mass of MgO formed in the reaction is

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21 grams

10 grams

42 grams

28 grams

answer is D.

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Detailed Solution

$\large \begin{array}{l} \mathop {2Mg}\limits_{(2 \times 24g)} {\rm{ }} + {\rm{ }}\mathop {{O_2}}\limits_{32g} \to {\rm{ }}\mathop {2MgO}\limits_{(2 \times 40g)} \\ 6gm\,\,\,\,\,\,\,\,\,\,\,\,\,?\,\,\,\,\left( { = 4gm} \right)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6gm\,\left( {given} \right) \end{array}$
2gm excess of O₂ is given
Mg ....limiting reagent
O₂ ....Excess reagent
(2 x 24g) of Mg given (2 x 40g) of MgO
6g of Mg given 'X' gram of MgO
$\large X\; = \;\frac{{6 \times 2 \times 40}}{{2 \times 24}} = 10g$