1) Consider the combustion reaction of methane, CH4: CH4(g) + 2O2 (g) ! CO2(g) +

Question

1)

Consider the combustion reaction of methane, CH4:
CH4(g) + 2O2 (g) ! CO2(g) + 2 H2O(l)
Suppose 8.00 g of CH4 is reacting with 8.00 g of oxygen. How much (in gram) CO2 and H2O are
produced and how much (in gram) excess reactant is left?

2)

Copper wire reacts with concentrated nitric acid aqueous solution to generate water soluble
copper(II) nitrate, nitrogen dioxide gas and water.
a. Write a balanced equation for the reaction.
b. If 6.61 g of Cu(NO3)2 are obtained from 4.55 g of Cu to react with 1.00 mol of HNO3, what is
the percent yield of the reaction?

Explanation / Answer

CH4(g) + 2O2 (g) ------------- CO2(g) + 2 H2O(l)

Molar mass of CH4= 16

Molar mass of O2 = 32

Molar mass of CO2= 44

Molar mass of H2O = 18

8.00 g of CH4 is equal to 8/16= 0.5 moles

8.00 g of oxygen is equal to 8/32 = 0.25Moles

Limiting reagent is oxygen

One equivalent of CH4 will consume of 2 equavalent of O2

which means 0.125 Mole of CH4 will react with 0.25 mole of O2 and to give 0.125 moles of CO2 (0.125 x 44= 5.5gm) and 0.25 Moles of H2O (0..25 x 18 = 4.5 gm)

rest 0.375 Moles of CH4 will be rest = 0.375 x 16 = 6 gm of CH4 will be unreacted

5.5gm of CO2 and 4.5 gm of H2O will be produced.

Cu(s) + 4HNO3(aq)

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