1.) C 3 H 8 + O 2 -> CO 2 + H 2 O How many grams of CO 2 can be produced by burn

Question

1.) C3H8 + O2 -> CO2 + H2O

How many grams of CO2 can be produced by burning 3.65 moles of propane? Assume oxygen is the excess reagent in this reaction.

2.) 6Li + N2 -> 2Li3N

What is the theoretical yield of Li3N in grams when 12.3g of Li are heated with 33.6g of N2? If the actual yield of Li3N is 5.89g, what is the percent yield of the reaction?

3.) FeTiO3 + H2SO4 -> TiO2 + FeSO4 + H2O

In one process 8.00x103kg of FeTiO3 yielded 3.67 x 103kg of TiO2. What is the percent yield of the reaction?

Explanation / Answer

1) C3H8 + 5O2 -> 3CO2 + 4H2O

as per the balanced reaction , 1 mole of propane produces 3 moles of CO2

thus, 3 moles of propane produces 9 moles of CO2

Molar mass of CO2 = 44 g/mole

thus, mass of CO2 produced = 9*44 = 396 g

2) 6Li + N2 -------> 2Li3N

As per the balanced reaction Li and N2 reacts in the ratio of 6:1

now, molar mass of Li = 7 g/mole

thus, moles of Li in 12.3 g of it = 12.3/7 = 1.757

molar mass of N2 = 28g/mole

moles of N2 in 33.6 g of  it = 33.6/28 = 1.2

Clearly Li is the limiting reagent

Thus, moles of Li3N produced from 1.757 moles of Li = 1.757/3 = 0.586 moles

molar mass of Li3N = 35 g/mole

hence mass of Li3N produced = 0.586*35 = 20.51 g

now, actual yield = 5.89 g

Thus, %yield = (actual yield/theoretical yield)*100 = (5.89/20.51)*100 = 28.72%

3)  FeTiO3 + H2SO4 -> TiO2 + FeSO4 + H2O

Molar mass of FeTiO3 is 151.7102 g/mol

molar mass of H2SO4 = 98 g/mole

molar mass of TiO2 =  79.87 g/mole

now, moles of FeTiO3 in 8000 kg of it = 8000/151.7102 = 52.73 moles

moles of TiO2 in 3670 g of it = 3670/79.87 = 45.95 moles

now, as per the balanced reaction , 1 mole of FeTiO3 must produce 1 mole of TiO2 theoretically

so, 52.73 moles of FeTiO3 must produce 52.73 moles of TiO2 theoretically

But actual TiO2 produced = 45.95 moles

thus, %yield = (moles of actual TiO2 produced/moles of TiO2 produced theoretically)*100 = (45.95/52.73)*100 = 87.14%

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