Question

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

How many grams of carbon should be burned in an excess of oxygen at STP to obtain 2.21 L of carbon dioxide?

1.18 g

2.21 g

4.12 g

4.34 g

Answer 1

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

Step-by-step 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 =
`(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
`(1)/(1)* 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* 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.

Step-by-step 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 =
`(1)/(22.4L)* 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 =
`(1)/(1)* 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.