Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 23. g of butane is mixed with 29.1 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

12.6 g of H₂O can be produced in the combustion

Explanation:

This excersise involves an easy combustion reaction:

2C₄H₁₀  +  13O₂  →   8CO₂  +  10H₂O

First step: calculate moles of reactants in order to find out the limiting.

23 g . 1mol / 58g = 0.396 moles of butane

29.1 g . 1mol /32g = 0.909 moles of O₂

2 moles of butane react to 13 moles of oxygen

Then, 0.396 moles of butane may react to (0.396 . 13) / 2 = 2.574 moles

Certainly we do not have enough oxygen, so O₂ becomes the limiting reactant

13 moles of O₂ can produce 10 moles of water

Then 0.909 moles may produce (0.909 . 10) /13 = 0.699 moles H₂O

We convert moles to mass → 0.699 mol . 18 g/mol = 12.6 g

hi,

the answer is d, it has indefinite shape but definite volume.

hope this !

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