Balanced equation of combustion of methane gas is shown below CH_4(g) + 2O_2righ

Question

Balanced equation of combustion of methane gas is shown below CH_4(g) + 2O_2rightarrow CO_2+, 2H_2O In the above balanced equation 4.12 mole of oxygen reacts with 3.51 moles CH_4. Calculate the limiting reactant? Show setup with conversion factors and finish with c v and fx. If you do not use correct conversion factors, points will be deducted. In a lab a student weighed 1.125 g of Cu metal and reacted with cone. HNO_3 in the fume hood. The mass of Cu(NO_3)_2 produced was 1.675 g. Calculate the percent yield of Cu in Cu(NO_3)_2. Show all calculation steps by using conversion factors. Also show molar mass calculation of Cu(NO_3)_2 to get full credit.

Explanation / Answer

As per the given balanced equation

CH4 + 2O2 --> CO2 + 2H2O

one mole of methane will react with two moles of O2

So the mole ration of methane to O2 should be 2:1

However, here the given ratio is =Methane : O2 = 3.51 / 4.12 < 2:1

So the limiting reagent is methane.

4.12 moles of Oxygen will react with 2.06 moles of methane

the heat of combustion for methane = sum of Delta H of formation of products - Sum of delta H of formation of reactants

Heat of combustion = [Delta Hformation of CO2 + 2X Delta H formation of H2O] - [ Delta H of foramtion of methane]

Heat of combusiton = [ -393.5 kJ/mol + 2 X -241.8kJ/mol] - [74.87 kJ/mol] = -802.23 KJ / mole

Calorific value = Heat of combustion / molecular weight = -802.23 / 16 = -50.14 KJ / gram

Fuel value = Heat of combusiton X moles of fuel present = -802.23 X 2.06 = 1652.59 KJ

2) The balanced equation will be

Cu(s) + 4HNO3(l) --> Cu(NO3)2 + 2 NO2(g) + 2 H2O(l)

so, one mole of copper (63.5 grams / mole= Molecular weight ) will react with four moles of HNO3 to give one mole of copper nitrate (187.56 g/mol)

so 63.5 grams will give 187.56 grams of copper nitrate

1 gram of copper will give = 187.56 / 63.5 grams of copper nitrate

1.125 grams of copper will give= 1.125 x 187.56 / 63.5 grams of copper nitrate = 3.32 grams of copper nitrate

So theoretical yield = 3.32 grams

Actual yield = 1.675 grams

% of yield = 1.675 X 100 / 3.32 = 50.45%

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