Question

1. How many moles of Al2O3 are formed from the reaction of 6.38grams of O2 and 9.15grams of Al?

2. How many grams of Al2O3 are formed from the reaction of 1.88grams of Al and 1.88 grams of O2?

3. How many grams of Al2O3 are formed from the reaction of 8.32x10^20 molecules of O2 and 4.26x10^21 molecules of Al?

Answer 1

4 Al + 3 O2 ------------------> 2 Al2O3

Al molar mass = 26.98 g /mol

O2 molar mass = 32 g /mol

Al2O3 molar mass = 101.96 g /mol

no of moles = mass / molar mass

1) moles of o2 = 6.38 / 32 = 0.199375 moles

                    Al = 9.15 / 26.98 = 0.34 moles

4 moles Al ---------------------> 2 moles Al2O3

0.34 moles ------------------>?

=> 0.34*2/4 = 0.169 moles

3 moles O2 --------------------> 2 Al2O3

0.199375 moles --------------->?

=> 0.199375 * 2 / 3 = 0.133 moles

limitting reactant is O2

mass of Al2O3 = 0.133 * 101.96 = 13.55 grams

2)

moles of o2 = 1.88 / 32 = 0.05875 moles

                    Al =1.88 / 26.98 = 0.06968 moles

4 moles Al ---------------------> 2 moles Al2O3

0.06968 moles ------------------>?

=> 0.06968*2/4 = 0.03484moles

3 moles O2 --------------------> 2 Al2O3

0.05875 moles --------------->?

=> 0.05875* 2 / 3 = 0.0392moles

limitting reactant is Al

mass of Al2O3 = 0.03484 * 101.96 = 3.55 grams

3) 1 mole ------------------> 6.02*10²³ molecules

for O2 = 8.32*10²⁰ / 6.02*10²³ = 0.00138 moles

for Al = 4.26*10²¹ / 6.02*10²³ = 0.007076 moles

4 moles Al ---------------------> 2 moles Al2O3

0.007076 moles ------------------>?

=> 0.007076*2/4 = 0.003538 moles

3 moles O2 --------------------> 2 Al2O3

0.00138 moles --------------->?

=> 0.00138 * 2 / 3 = 0.000921 moles

limitting reactant is O2

mass of Al2O3 =0.000921 * 101.96 = 0.094 grams