Question

What is the maximum mass (in g) of NO, formed when 7.5 g of N20 are consumed? 2N205(g) → 4N02(g) + 02(g)

Answer 1

To find the maximum mass of NO formed when 7.5 g of N2O5 is consumed, we need to determine the limiting reactant in the reaction with NO2.

Step-by-step explanation:

To find the maximum mass of NO formed, we need to determine the limiting reactant in the reaction between N2O5 and NO2. First, we need to calculate the molar masses of N2O5 and NO2, which are 108 g/mol and 46 g/mol respectively. Next, we can convert the given mass of N2O5 (7.5 g) to moles using its molar mass. Then, using the stoichiometry of the reaction, we can determine the moles of NO produced. Finally, we can convert the moles of NO to grams using its molar mass to find the maximum mass of NO formed.

Given:
Mass of N2O5 = 7.5 g
Molar mass of N2O5 = 108 g/mol
Molar mass of NO2 = 46 g/mol

First, calculate the moles of N2O5:
moles of N2O5 = mass of N2O5 / molar mass of N2O5
moles of N2O5 = 7.5 g / 108 g/mol

= 0.0694 mol

Since the stoichiometry of the reaction is 2N2O5 : 4NO2, the moles of NO produced will be twice the moles of N2O5:
moles of NO = 2 x moles of N2O5

= 2 x 0.0694 mol

= 0.1388 mol

Finally, convert the moles of NO to grams:
mass of NO = moles of NO x molar mass of NO
mass of NO = 0.1388 mol x 46 g/mol

= 6.39 g