Answer to Question #300472 in General Chemistry for Jane

Question #300472

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



1
Expert's answer
2022-02-22T06:37:02-0500

Molar mass of C3H8=44.1C_3H_8=44.1

Molar mass of O2=32.0_2=32.0

Molar mass of H2O=18.0_2O=18.0


Moles of C3H8 =14.844.1=0.3356=\frac{14.8}{44.1}=0.3356


Moles of O2=3.4432=0.1075_2=\frac{3.44}{32}=0.1075


Grams of H2O produced =0.1075×18=0.1075×18

=1.935g=1.935g


Limiting reagent is O2


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


Excess of C3H8 =14.84.741=14.8-4.741

=10.059g=10.059g





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