Question

C3H8+5O2 ---> 3CO2+4H2O
If you start with 14.8 g of C3
H8 and 3.44 g of O2, determine the limiting reagent
c) determine the number of grams of H2
O produced
d) determine the number of grams of excess reagent left

Answer 1

for one mole of C3 H8 u need 5 mole of oxygen
that is 44 gram of C3H8 needs 160 gram of O2
for 14.8 gram of C3H8 u need =160*14.8 /44
= 53.81 gram of Oxygen bt you have only 3.44 gram hence O2 is limiting agent

Answer 2

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