Final answer:
To determine how many grams of butane can be burned by 1.42 moles of oxygen, we use the balanced chemical equation for the combustion of butane and set up a proportion to find the mass of butane. The answer is 12.73 grams.
Step-by-step explanation:
To determine how many grams of butane can be burned by 1.42 moles of oxygen, we need to use the balanced chemical equation for the combustion of butane, which is:
C4H10 + 13/2 O2 ⟶ 4 CO2 + 5 H2O
From the balanced equation, we can see that 13/2 moles of oxygen are required to burn 1 mole of butane. Therefore, to find out how many grams of butane can be burned by 1.42 moles of oxygen, we can set up a proportion:
(1.42 moles O2) / (13/2 moles O2) = (x grams butane) / (1 mole butane)
Solving for x, we find that x is equal to (1.42 moles O2) * (1 mole butane) / (13/2 moles O2) = 0.2192 moles butane.
Since the molar mass of butane (C4H10) is 58.12 g/mol, we can calculate the mass of butane:
0.2192 moles butane * 58.12 g/mol = 12.73 grams of butane can be burned by 1.42 moles of oxygen.