Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 0.58 g of butane is mixed with 0.874 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.

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

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