Balanced equation of combustion of methane gas is shown below. CH_4(g) + 2O_2 ri

Question

Balanced equation of combustion of methane gas is shown below. CH_4(g) + 2O_2 rightarrow 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 f.v. 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 conc. 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 pe the reaction CH4+2O2->CO2 + 2H2O

Molar ratio of CH4: O2= 1:2

Actual ratio = 3.51: 4.12 = 1:1.17

Since the ratio of oxygen is less in actual case, oxygen is the limiting reactant

2.

Atomic weight of Cu= 63.5, moles of copper= mass/atomic weight= 1.25/63.5=0.01717

3Cu+8 HNO3 => 3Cu(NO3)2 + 2 NO+ 4H2O

3 moles of Cu gives 3 moles of Cu(NO3)2

0.01717 moles gives 0.017717 moles when 100% conversion takes place (also known as theoretical yield)

Molar mass of Cu(NO3)2= 188

Mass of Cu(NO3)2 =moles*molar mass= 0.017817*188= 3.33

Mass of Cu(NO3)2 obtained = 1.675

% yield = 100*1.675/3.33 = 50.3%

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