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1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24, O = 16)

(a) Mg, 0.16 g

(c) Mg, 0.44 g

(b) O2, 0.16 g

(d) O2, 0.28 g

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Answer - (a) Mg, 0.16g

 

Solution - 

2Mg + O2 → 2MgO

no of moles of Mg = 1/24 = 0.0416 moles

no of moles of O2 = 0.56/32 = 0.0175 moles 

0.0416/2 = 0.0208 > 0.0175 

thus O2 is limiting reagent 

ReactantsMgO2
Initial moles0.04160.0175
Final moles0.0416 - (2 x0.0175) = 0.0066 0.0175 - (1 x 0.0175) = 1


mass of Mg = 0.0066 x 24 = 0.16g 

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