Answer to Question #300472 in General Chemistry for Jane

Question

C3H8 + O2                       CO2 + H2O

If you start with 14.8g of C3H8 and 3.44g of O2, how many grams of H2O are produced?

Which is the limiting reagent (which reactant created less H2O? _______

How many grams of the non-limiting reagent are actually used?

How many grams of the non-limiting reagent are in excess?

Note : Balance first the equation

It becomes    C3H8 +5O2              3 CO2 + 4H2O

Answer 1

Molar mass of $C_3H_8=44.1$

Molar mass of O$_2=32.0$

Molar mass of H$_2O=18.0$

Moles of C₃H₈ $=\frac{14.8}{44.1}=0.3356$

Moles of O$_2=\frac{3.44}{32}=0.1075$

Grams of H₂O produced $=0.1075×18$

$=1.935g$

Limiting reagent is O₂

Grams of C₃H₈ used $=0.1075×44.1=4.741g$

Excess of C₃H₈ $=14.8-4.741$

$=10.059g$