How many moles of water vapor could be produced by the combustion of 3.50 moles of
propane (C3H8) according to the following equation: C3H8 + 5 O2 --> 3 CO2 + 4H20

A) 14.00
B) 17.50
C) .88
D) 10.50

A. 14.00

Explanation:

Looking at the given balanced equation for the combustion of Propane

C3H8  +  5 O2 > 3 CO2  +  4H20

1 mole of Propane produces 4 moles of water vapor.

∴ 3.50 moles of  Propane would have an equation such as this, when the coefficients is multiplied by 3.50

3.5 C3H8  + 17.5 O2 > 10.5 CO2  + 14 H2O

This imply that 14 moles of water vapor is produced by the combustion of 3.50 moles of Propane

the temperature of the gas, in k, not celsius.