Chemistry, 18.02.2020 04:19 jtbrown0093

A sample of 8.00 g of liquid 1‑propanol, C 3 H 8 O , is combusted with 49.3 g of oxygen gas. Carbon dioxide and water are the products. Write the balanced chemical equation for the reaction. Physical states are optional. chemical reaction: What is the limiting reactant? oxygen 1‑propanol How many grams of CO 2 are released in the reaction? mass of CO 2 : g How many grams of the excess reactant remain after the reaction is complete? mass of excess reactant remaining:

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Answer from: 200002977

The limiting reagent is 1-propanol and the excess reagent is oxygen gas and its mass remaining is 30.144 grams. The mass of carbon dioxide released is 17.56 grams

Explanation:

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

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

For 1‑propanol:

Given mass of 1‑propanol= 8.00 g

Molar mass of 1‑propanol= 60g/mol

Putting values in equation 1, we get:

$\text{Moles of 1‑propanol}=\frac{8.00g}{60g/mol}=0.133mol$

For oxygen gas:

Given mass of oxygen gas = 49.3 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{49.3g}{32g/mol}=1.54mol$

The chemical equation for the combustion of 1-propanol follows:

$2C_3H_8O+9O_2\rightarrow 6CO_2+8H_2O$

By Stoichiometry of the reaction:

2 moles of 1‑propanol reacts with 9 moles of oxygen gas

So, 0.133 moles of 1‑propanol will react with = $\frac{9}{2}\times 0.133=0.598mol$ of oxygen gas

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

Thus, 1‑propanol is considered as a limiting reagent because it limits the formation of product.

Moles of excess reagent (oxygen gas) = [1.54 - 0.598] = 0.942 moles

By Stoichiometry of the reaction:

2 moles of 1‑propanol produces 6 moles of carbon dioxide

So, 0.133 moles of 1‑propanol will produce = $\frac{6}{2}\times 0.133=0.399moles$ of carbon dioxide.

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

For oxygen gas:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 0.942 moles

Putting values in equation 1, we get:

$0.942mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.942mol\times 32g/mol)=30.144g$

For carbon dioxide gas:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.399 moles

Putting values in equation 1, we get:

$0.399mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.399mol\times 44g/mol)=17.56g$

Hence, the limiting reagent is 1-propanol and the excess reagent is oxygen gas and its mass remaining is 30.144 grams. The mass of carbon dioxide released is 17.56 grams

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A sample of 8.00 g of liquid 1‑propanol, C 3 H 8 O , is combusted with 49.3 g of oxygen gas. Carbon...

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