2C4H10+1302 → 8CO2 + 10H2Oassuming oxygen is the limiting reagent, how much butane would remain if 20g grams butane burns in 60 g of oxygen

First, we need to determine the molar mass of the reactants and products in the chemical equation.

The molar mass of butane (C4H10) is approximately 58.12 g/mol. The molar mass of oxygen (O2) is approximately 32 g/mol.

Next, we need to convert the given mass of the reactants into moles.

For butane: 20 g ÷ 58.12 g/mol = 0.344 moles For oxygen: 60 g ÷ 32 g/mol = 1.875 moles

The stoichiometry of the reaction tells us that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.

Therefore, the mole ratio of butane to oxygen is 2:13.

To find out how much butane would react with the given amount of oxygen, we can set up a proportion:

2/13 = x/1.875

Solving for x gives us approximately 0.288 moles of butane.

Since we started with 0.344 moles of butane, the amount of butane that would remain after the reaction is:

0.344 moles - 0.288 moles = 0.056 moles

Finally, we convert this amount back into grams using the molar mass of butane:

0.056 moles x 58.12 g/mol = approximately 3.25 g

So, approximately 3.25 g of butane would remain if 20 g of butane burns in 60 g of oxygen, assuming oxygen is the limiting reagent.