Question

For the following reaction, 9.37 grams of carbon (graphite) are allowed to react with 12.6 grams of oxygen gas . carbon (graphite)(s) + oxygen(g) carbon dioxide(g) What is the maximum mass of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams Submit AnswerRetry Entire Group

Answer 1

Answer: The maximum mass of carbon dioxide formed is 17.34 grams, the formula of limiting reagent is
`O_2` and the mass of excess reagent (carbon) remained is 4.644 grams

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For carbon:

Given mass of carbon = 9.37 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

`\text{Moles of carbon}=(9.37g)/(12g/mol)=0.781mol`

• For oxygen gas:

Given mass of oxygen gas = 12.6 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(12.6g)/(32g/mol)=0.394mol`

The chemical equation for the reaction of carbon and oxygen gas follows:

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

By Stoichiometry of the reaction:

1 moles of oxygen gas reacts with 1 mole of carbon metal.

So, 0.394 moles of oxygen gas will react with =
`(1)/(1)* 0.394=0.394mol` of carbon metal.

As, given amount of carbon metal is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (carbon) left = [0.781 - 0.394] = 0.387 moles

By Stoichiometry of the reaction:

1 moles of oxygen gas produces 1 mole of carbon dioxide

So, 0.394 moles of oxygen gas will produce =
`(1)/(1)* 0.394=0.394moles` of carbon dioxide

Now, calculating the mass of carbon and carbon dioxide from equation 1, we get:

• For carbon:

Excess moles of carbon = 0.387 moles

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

`0.387mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=(0.387mol* 12g/mol)=4.644g`

• For carbon dioxide:

Moles of carbon dioxide = 0.394 moles

Molar mass of carbon dioxide = 44 g/mol

Putting values in equation 1, we get:

`0.394mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.394mol* 4g/mol)=17.34g`

Hence, the maximum mass of carbon dioxide formed is 17.34 grams, the formula of limiting reagent is
`O_2` and the mass of excess reagent (carbon) remained is 4.644 grams