Chemistry stoichiometry

Question

calculate how many moles of NO2 form when each quantity of reactant completely reacts

2N2O5 (g) -> 4NO2 (g) + O2 (g)

a.) 2.5 mol N2O5

b.) 6.8 mol N2O5

c.) 15.2 g N2O5

d.) 2.87 kg N2O5

Answer 1

equation balanced - 2N2O5 (g)-------> 4NO2 (g) + O2 (g)

a) starting with 2.5 mol N2O5 as reactant to determine # moles of NO2 product forms

Using.the molar ratio of N2O5 to NO2

2.5 mol N2O5 X 4 mol NO2/2 mol N2O5 = 5 mol NO2

b) starting with 6.8 mol N2O5 again using the molar ratios of N2O5 to NO2

6.8 mol N205 x 4 mol NO2/2 mol N2O5 = 13.6 mol NO2

c) starting with 15.2 g of N2O5

Convert grams of N2O5 into moles

first calculate the Molar Mass of N2O5 - N 2 x 14g = 28g

----------------------------------------------------- O 5 x 16g = 80g

------------------------------------- Total = ------------------------108g the molar mass

15.2 g N2O5 x 1 mol N2O5/108g N2O5 = 0.14 mol N2O5

Now use the molar ratio of N2O5 to NO2

0.14 mol N2O5 x 4 mol NO2/2 mol N2O5 = 0.28 mol NO2

d) starting with 2.870 Kg of N2O5

convert Kg N2O5 to grams = 2870g N2O5

convert grams of N2O5 to mol of N2O5

2870g N2O5 x 1 mol N2O5/108g N2O5 = 26.57 mol N2O5

Use the molar ratio of N2O5 to NO2

26.57 mol N2O5 x 4 mol NO2/2 mol N2O5 = 53.14 mol NO2