The question:A new grill has a mass of 30.0 kg. You put 2.0 kg of charcoal in the grill. You burn all the charcoal and the grill has a mass of 30.0 kg. What is the mass in kg of the gases given off? (assume that the charcoal is pure carbon solid and that it burns completely in oxygen).

The answer: Mass of carbon burnt = 2Kg = 2000g
Moles of Carbon burnt = 2000/12
=> Moles of CO2 produced = 2000/12
=> Mass of CO2 produced = 2000/12*44 = 2 333.33 g = 2.33 Kg

In the answer, where did the 44 come from?

Question

01-10-2018
Chemistry

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The question:A new grill has a mass of 30.0 kg. You put 2.0 kg of charcoal in the grill. You burn all the charcoal and the grill has a mass of 30.0 kg. What is the mass in kg of the gases given off? (assume that the charcoal is pure carbon solid and that it burns completely in oxygen).

The answer: Mass of carbon burnt = 2Kg = 2000g
Moles of Carbon burnt = 2000/12
=> Moles of CO2 produced = 2000/12
=> Mass of CO2 produced = 2000/12*44 = 2 333.33 g = 2.33 Kg

In the answer, where did the 44 come from?

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tallinn tallinn · Answer 1 · 2018-10-10T05:20:39+02:00

Answer:- Molar mass of [tex]CO_2[/tex] .

Solution:- It is a stoichiometry problem. Mass of the grill is 30.0 kg and the mass after burning the grill is also 30.0 kg. It means all the charcol is burned and the gas is given off.

2.0 kg of charcol are converted to grams which is 2000 g. Since charcol is pure solid carbon, the grams are divided by the atomic mass of carbon which is 12.

The combustion equation of charcol is written as:

[tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]

From this balanced equation, there is 1:1 mol ratio between charcol and carbon dioxide. So, the moles of carbon dioxide gas formed are equivalent to the moles of charcol. To convert the moles of carbon dioxide to grams we multiply the moles by it's molar mass.

Carbon dioxide has one carbon and two oxygen atoms so it's molar mass = 12 + 2(16) = 12 + 32 = 44

So, 44 is the molar mass of carbon dioxide and above calculations clearly shows how and where we get this.