4. For the reaction Barn methane (CH4) if 28.6 grams of oxygen how much carbon d

Question

4. For the reaction Barn methane (CH4) if 28.6 grams of oxygen how much carbon dioside could be produced? grams of methane reacts with 57.6 5. What is the percentage yield of carbon dioxide in reaction described by question 6 if 32.1 grams of carbon dioxide are actually produce 6. In the many moles of excess reactant are left at the end of the reaction? reaction described by question 4 which reactant is the limiting reactant and how 7. A 29.1 gram piece of lead absorbed 38.4 calories of heat energy when it was heated from 23 to 64°C. How much heat would be required to raise 0.0291 kg of lead by the same temperature difference? a. 0.0384 cal b. 3.84 cal c. 38.4 cal d. 38.4 kcal

Explanation / Answer

4)

Molar mass of CH4 = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass of CH4 = 228.6 g

we have below equation to be used:

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(228.6 g)/(16.042 g/mol)

= 14.25 mol

Molar mass of O2 = 32 g/mol

mass of O2 = 57.6 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(57.6 g)/(32 g/mol)

= 1.8 mol

we have the Balanced chemical equation as:

CH4 + 2 O2 ---> CO2 + 2 H2O

1 mol of CH4 reacts with 2 mol of O2

for 14.2501 mol of CH4, 28.5002 mol of O2 is required

But we have 1.8 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

From balanced chemical reaction, we see that

when 2 mol of O2 reacts, 1 mol of CO2 is formed

mol of CO2 formed = (1/2)* moles of O2

= (1/2)*1.8

= 0.9 mol

we have below equation to be used:

mass of CO2 = number of mol * molar mass

= 0.9*44.01

= 39.6 g

Answer: 39.6 g

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