If 62.0 grams of magnesium metal (Mg) react with 55.5 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the excess reactant will be left over when the reaction is complete? Be sure to write out the balanced equation for this reaction and to show all of your work.

Answers

Answer 1

Answer: The balanced chemical equation would be as follows:

Mg + O2 → MgO2

We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:

62.0 g Mg (1 mol / 24.31 g ) = 2.55 mol Mg
55.5 g O2 ( 1 mol / 32 g ) = 1.74 mol O2 -----> consumed completely and therefore the limiting reactant

2.55 mol - 1.74 mol O2( 1 mol Mg / 1 mol O2 ) = 0.81 mol Mg excess

Answer 2

Answer:

Answer:

14.1648 grams of oxygen gas will be left.

Explanation:

$2Mg +O_2\rightarrow 2MgO$

Moles of magnesium metal = $(62.0 g)/(24 g/mol)=2.5833 mol$

Moles of oxygen gas = $(55.5 g)/(32 g/mol)=1.7343 mol$

According to reaction, 2 mol of magnesium react with 1 mol of oxygen gas .

Then 2.5833 moles of magnesium will react with:

$(1)/(2)* 2.5833 mol=1.29165 mol$ of oxygen gas.

Moles of oxygen left unreacted =1.7343 mol - 1.29165 mol = 0.44265 mol

Oxygen gas is an excessive reagent.

Mass of 0.44265 moles of oxygen gas:

0.44265 mol × 32 g/mol = 14.1648 g

14.1648 grams of oxygen gas will be left.

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Answers

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