Suppose a student reacts 0.3470 g Mg with O2 gas. In his/her calculations, however, s/he mistakenly...

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Suppose a student reacts 0.3470 g Mg with O2 gas. In his/her calculations, however, s/he mistakenly...

Suppose a student reacts 0.3470 g Mg with O₂ gas. In his/her calculations, however, s/he mistakenly uses 0.2470 g Mg. Will the theoretical yield of MgO be too high, too low, or unaffected?

science chemistry

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Answer 1

Answer #1

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The reaction is given as:

2Mg + O2 = 2MgO

The actual yield of MgO is directly related to the amount of Mg used , so it will be obtained according to the mass of Mg used in the reaction i.e. 0.3470 g and the percent yield of reaction.

However if in the calculations , the mass of Mg used by the student is 0.2470 g , the theoretical yield will also come to be too low.

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Answer 2

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Suppose a student reacts 0.3470 g Mg with O2 gas. In his/her calculations, however, s/he mistakenly...

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For the following reaction, 2Mg(s) + O2(g) → 2MgO (s), when 10.1 g of Mg reacts with 10.5 g of O2

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Suppose a student reacts 0.3470 g Mg with O2 gas. In his/her calculations, however, s/he mistakenly...

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Add Answer to: Suppose a student reacts 0.3470 g Mg with O2 gas. In his/her calculations, however, s/he mistakenly...

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Suppose a student reacts 0.3470 g Mg with O2 gas. In his/her calculations, however, s/he mistakenly...