Given the chemical equation, 2Mg + O2 —> 2MgO, when 2.2 g Mg react with 3.6 g of

Question

Given the chemical equation, 2Mg + O2 —>  2MgO, when 2.2 g Mg react with 3.6 g of O2, 2.7 g MgO were obtained. What is the percent yield in the reaction?
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Explanation / Answer

Molar mass of Mg = 24.31 g/mol

mass of Mg = 2.2 g

we have below equation to be used:

number of mol of Mg,

n = mass of Mg/molar mass of Mg

=(2.2 g)/(24.31 g/mol)

= 9.05*10^-2 mol

Molar mass of O2 = 32 g/mol

mass of O2 = 3.6 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(3.6 g)/(32 g/mol)

= 0.1125 mol

we have the Balanced chemical equation as:

2 Mg + O2 ---> 2 MgO +

2 mol of Mg reacts with 1 mol of O2

for 0.0905 mol of Mg, 0.0452 mol of O2 is required

But we have 0.1125 mol of O2

so, Mg is limiting reagent

we will use Mg in further calculation

Molar mass of MgO = 1*MM(Mg) + 1*MM(O)

= 1*24.31 + 1*16.0

= 40.31 g/mol

From balanced chemical reaction, we see that

when 2 mol of Mg reacts, 2 mol of MgO is formed

mol of MgO formed = (2/2)* moles of Mg

= (2/2)*0.0905

= 9.05*10^-2 mol

we have below equation to be used:

mass of MgO = number of mol * molar mass

= 9.05*10^-2*40.31

= 3.648 g

% yield = actual mass*100/theoretical mass

= 2.7*100/3.648

= 74 %

Answer: 74 %

Feel free to comment below if you have any doubts or if this answer do not work. I will correct it and submit again if you let me know

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