In the reaction $2 M g + O_{2} \to 2 M g O$ $2Mg + O_2 -> 2MgO$ , the law of definite proportions states that for every 2 moles of $M g$ $Mg$ you will need how many moles of $O_{2}$ $O_2$ ?

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In the reaction $2 M g + O_{2} \to 2 M g O$ $2Mg + O_2 -> 2MgO$ , the law of definite proportions states that for every 2 moles of $M g$ $Mg$ you will need how many moles of $O_{2}$ $O_2$ ?

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Answer 1 · 2016-02-28T11:43:13

1.

Explanation:

It's in the balanced equation. Each number before a compound corresponds to how many miles you use, and when there is no number you put in a 1. So 2 $M g$ $Mg$ + (1) $O_{2}$ $O_{2}$ means twice as many moles of $M g$ $Mg$ as $O_{2}$ $O_{2}$ .

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In the reaction $2 M g + O_{2} \to 2 M g O$ $2Mg + O_2 -> 2MgO$ , the law of definite proportions states that for every 2 moles of $M g$ $Mg$ you will need how many moles of $O_{2}$ $O_2$ ?

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In the reaction 2Mg+O2→2MgO2Mg + O_2 -> 2MgO, the law of definite proportions states that for every 2 moles of MgMg you will need how many moles of O2O_2?

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In the reaction $2 M g + O_{2} \to 2 M g O$ $2Mg + O_2 -> 2MgO$ , the law of definite proportions states that for every 2 moles of $M g$ $Mg$ you will need how many moles of $O_{2}$ $O_2$ ?