# Stoichiometry Problems in Chemistry

Stoichiometry problems in chemistry and how to solve them by the CHEMIX School program

Stoichiometry is about calculation of the quantities/relationship of/between reactants and products in a chemical reaction. It is about calculations of moles, masses and  percents within a chemical equation.

Stoichiometry problem/example No.1

A piece of metallic iron (10moles) was dissolved in conc HCl. The reaction formed H2 and FeCl2.

a) Balance: Fe+HCl>FeCl2+H2

and determine

b) amount of formed FeCl2

b) amount of used HCl

The solution

a)     Fe + 2HCl = FeCl2+H2

b)    As we can see from the balanced equation (a)  n(Fe) = n(FeCl2)  , 10 moles of FeCl2 was formed in the reaction

c) The balanced equation (a) tells us that n(HCl) = 2n(Fe) --> 20 moles of HCl was used.

The CHEMIX School Stoichiometry Problem solution

Start CHEMIX School and select push button Balance

Insert  10mFe + HCl > FeCl2 + H2  in the unbalanced equation text field and press push button Calculate Stoichiometry example/problem No.2

In following unbalanced reaction 10 grams pentane was burning:   C5H12 + O2 --> CO2 + H2O

a) Balance    C5H12 + O2 --> CO2 + H2O

b) How many grams of oxygen was used ?

c) How many grams of water was formed ?

The solution

a)  C5H12 + 8O2 = 5CO2 + 6H2O

b) 10 grams of pentane = 0.1386 moles   n(O2) = 8n(C5H12)

8 times this amount (0.1386 moles C5H12) of oxygen was used --> 8*0.1386mole*32g/mole = 35.48grams O2

c) From (a) we know that 6 times more moles of water was formed than used pentane (0.1386 moles)

n(H2O) = 6n(C5H12)      -->   6*0.1386mole*18g/mole = 14.98 grams of water

The CHEMIX School Stoichiometry Problem solution

Start CHEMIX School and select push button Balance

Insert  10gC5H12 + O2 > CO2 + H2O   in the unbalanced equation text field and press push button Calculate

How to download CHEMIX School for solving Stoichiometry Problems

You can download CHEMIX School and solve stoichiometry problems by clicking the download link at the top of this page. Try it.