1) Combining 0.269 mol of Fe2O3 with excess carbon produced 10.3 g of Fe. Fe2O3

ID: 795721 • Letter: 1

Question

1) Combining 0.269 mol of Fe2O3 with excess carbon produced 10.3 g of Fe.

Fe2O3 + 3C ---> 2 Fe + 3CO

What is the actual and theoretical yield of iron in moles?

What was the percent yield?

2) Assuming an efficiency of 42.00%, calculate the actual yield of magnesium nitrate formed from 117.8 g of magnesium and excess copper(II) nitrate.

Mg+ Cu(NO3)2 ---> Mg(NO3)2 +Cu

3) Complete combustion of 4.40 g of a hydrocarbon produced 14.1 g of CO2 and 5.04 g of H2O. What is the empirical formula for the hydrocarbon?

Fe2O3 + 3C ---> 2 Fe + 3CO

1 mol of Fe2O3 with excess carbon produced 2mol of Fe.

0.269 mol of Fe2O3 with excess carbon produced 0.269*2mol=0.538mol of Fe.

but Fe produced is=10.3g=10.3/56 mol=0.184mol

actual yield of iron in moles=0.184mol

theoretical yield of iron in moles=0.538

percent yield=(0.184/0.538)*100=34.19%

2) Mg+ Cu(NO3)2 ---> Mg(NO3)2 +Cu

117.8 g of magnesium= 117.8/24=4.908 mole

1mol of magnesium and excess copper(II) nitrate produce 1mol magnesium nitrate

4.908 mol of magnesium and excess copper(II) nitrate produce 4.908mol magnesium nitrate

magnesium nitrate molar mass=148.3g/mol

theoritical yield=4.908mol magnesium nitrate=4.908*148.3=727.90g

actual yield=727.90*0.42=305.72g

3) 4.40 g of a hydrocarbon produced

14.1 g of CO2 =(12/44)*14.1=3.845 g C =0.3204 mol C

5.04 g of H2O=(2/18)*5.04=0.56g H =0.56 mol of H

O present in the hydrocarbon=4.40-(0.3204+0.56)=3.5196g of O=0.219975 mol O

C : H : O = 0.3204 : 0.56 : 0.219975 =1.45 : 2.55 :1 = 3 :5 :2

formula C3H5O2

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