How to do it not sure i tried but not sure

Mass =  17.28 g

Explanation:

Given data:

Mass of propane = 35 g

Volume of oxygen = 27 L

Mass of water produced = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂  →  3CO₂ + 4H₂O

First of all we will calculate the number of moles of oxygen at STP.

PV = nRT

n = PV/RT

n = 1 atm×27 L / 0.0821 atm. L/ mol.K × 273 K

n = 27 atm . L / 22.413 atm. L/mol

n = 1.2 mol

Moles of propane:

Number of moles = mass/ molar mass

Number of moles = 35 g/ 44.1 g/mol

Number of moles = 0.8 mol

Now we will compare the moles of water with propane and oxygen.

                           C₃H₈         :           H₂O

                               1             :             4

                              0.8          :            4×0.8 = 3.2 mol

                             O₂            :             H₂O

                              5             :                4

                             1.2            :             4/5×1.2 = 0.96 mol

Number of moles of water produced by oxygen are less so it will limiting reactant.

Mass of water:

Mass = number of moles × molar mass

Mass = 0.96 mol × 18 g/mol

Mass =  17.28 g

the answer is d. 74.02

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