Question

C(S)+O2(g)-->CO2(g)

Answer 1

Answer: The mass of carbon burned in oxygen at STP is 1.18 grams.

Explanation:

At STP:

22.4 L of volume is occupied by 1 mole of gas

We are given:

Volume of carbon dioxide = 2.21 L

So, moles of carbon dioxide gas will be = $\frac{2.21}{22.4}=0.0987mol$

For the given chemical reaction:

$C(s)+O_2(g)\rightarrow CO_2(g)$

By Stoichiometry of the reaction:

1 mole of carbon dioxide gas is formed when 1 mole of carbon is reacted.

So, 0.0987 moles of carbon dioxide is formed when $\frac{1}{1}\times 0.0987=0.0987mol$ of carbon id reacted.

Now, calculating the mass of carbon by using equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of carbon = 12 g/mol

Moles of carbon = 0.0987 moles

Putting values in above equation, we get:

$0.0987mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=(0.0987mol\times 12g/mol)=1.18g$

Hence, the mass of carbon burned in oxygen at STP is 1.18 grams.

Answer 2

Answer: The correct answer is 1.18 g.

Explanation:

We are given a chemical equation:

$C(S)+O2(g)\rightarrow CO_2(g)$

We know that at STP conditions:

22.4L of volume is occupied by 1 mole of a gas.

So, 2.21L of carbon dioxide is occupied by = $\frac{1}{22.4L}\times 2.21L=0.0986mol$ of carbon dioxide gas.

By Stoichiometry of the above reaction:

1 mole of carbon dioxide gas is produced by 1 mole of carbon

So, 0.0986 moles of carbon dioxide is produced by = $\frac{1}{1}\times 0.0986=0.0986mol$ of carbon.

Now, to calculate the mass of carbon, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of carbon = 0.0986 mol

Molar mass of carbon = 12 g/mol

Putting values in above equation, we get:

$0.0986mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=1.18g$

Hence, the correct answer is 1.18 g.