Asked by vera
If 120. g of propane, C3H8, is burned in excess oxygen according to the following equation: C3H8 + 5O2 3CO2 + 4H2O, how many grams of water are formed?
Answers
Answered by
MathMate
Molecular mass of propane (C3H8)
= 3*12+8*1
= 44
Molecular mass of water (H2O)
= 2*1+1*16
= 18
using the balanced equation, we find that the ratio
mass of propane : mass of water
= 1*44 : 4 * 18
= 44 : 72
Hence the mass of water can be calculated.
Note: replace the above approximate atomic masses by the exact values and repeat above calculations.
= 3*12+8*1
= 44
Molecular mass of water (H2O)
= 2*1+1*16
= 18
using the balanced equation, we find that the ratio
mass of propane : mass of water
= 1*44 : 4 * 18
= 44 : 72
Hence the mass of water can be calculated.
Note: replace the above approximate atomic masses by the exact values and repeat above calculations.
Answered by
Blessings
196.36g
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