When 91.5 g of isopropyl alcohol which has an empirical formula of C3H8O is burned in excess oxygen gas, how many grams of H2O are formed? MWC = 12.011 g/mol, MWH = 1.00794 g/mol, and MWO = 15.9994 g/mol.

109.7178g of H2O

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

2C3H8O + 9O2 —> 6CO2 + 8H2O

Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:

Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.

Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g

Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol

Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g

From the equation,

120.19184g of C3H8O produced 144.12224g of H20.

Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O

i believe the answer is three water molecules