37.4 g sample iron ore Fe2O3 mass 12.5 g

37.4 g sample iron ore Fe2O3 mass 12.5 g

 

 

A 37.4 g sample of iron ore is treated as follows.

The iron in the sample is all converted by

a series of chemical reactions to Fe2O3.

The mass of Fe2O3 is measured to be 12.5 g.

What was the mass of iron in the sample of ore?

What was the percent of iron in the sample of ore?

 

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> Fe2O3의 몰질량 = 159.69 g/mol

> Fe의 몰질량 = 55.85 g/mol

 

 

 

mass of Fe

= mass of Fe2O3 × [(2 × molar mass of Fe) / (molar mass of Fe2O3)]

= 12.5 × [(2 × 55.85) / (159.69)]

= 8.7435 g

 

 

 

(8.7435 g / 37.4 g) × 100 = 23.38%

---> percent of iron

 

 

 

답: 8.74 g, 23.4%