Answer:
0.378 g of H₂₂O can be produced.
Explanation:
The combustion reaction is:
2CH₃(CH₂)₂CH₃ + 13O₂ → 8CO₂ + 10H₂O
We convert the mass of reactants to moles:
0.58 g . 1mol / 58.1g = 0.00998 moles of butane
0.874 g . 1mol / 32g = 0.0273 moles of O₂
Oxygen is the limiting reactant. Look at stoichiometry.
2 moles of butane need 13 moles of oxygen to react
Then, 0.00998 moles of gas may react to (0.00998 . 13) / 2 = 0.06487 moles of oyxgen. I only have 0.0273 moles, so i do not have enough oxygen to complete the reaction.
Let's find out the product.
13 moles of oyxgen can produce 10 moles of water.
Then 0.0273 moles of O₂ may produce (0.0273 . 10)/13 = 0.021 moles
We convert to mass → 0.021 mol . 18g /1mol = 0.378 g
