<u>Answer:</u> 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
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of 1‑propanol= 8.00 g
Molar mass of 1‑propanol= 60g/mol
Putting values in equation 1, we get:
Given mass of oxygen gas = 49.3 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the combustion of 1-propanol follows:
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 = 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 = of carbon dioxide.
Now, calculating the mass of oxygen gas and carbon dioxide from equation 1, we get:
Molar mass of oxygen gas = 32 g/mol
Moles of oxygen gas = 0.942 moles
Putting values in equation 1, we get:
- <u>For carbon dioxide gas:</u>
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 0.399 moles
Putting values in equation 1, we get:
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