If 33.6 g of NO and 26.9 g of O2 react together, what is the mass in grams of NO2 that can be formed via the reaction below?

Question

2 NO (g) + O (g) → 2 NO (g)

Second Question: What volume of NO2 could be isolated at 945 mbar and 28 °C?

Answer 1

Write a correctly balanced equation:

2NO(g) + 0₂(g) ==> 2NO₂(g) ... balanced equation

Next, find which reactant is limiting:

moles NO present = 33.6 g NO x 1 mol NO/30 g = 1.12 moles NO present

moles O₂ present = 26.9 g O₂ x 1 mol O₂/32 g = 0.841 moles O₂ present

Since according to the balanced eq. it takes TWO moles NO for each ONE mol O₂, NO is limiting.

Next, based on the limiting reactant, calculate moles of NO₂ produced:

1.12 moles NO x 2 moles NO₂/2 moles NO = 1.12 moles NO₂ produced

Finally, convert moles NO₂ to grams NO₂:

1.12 moles NO₂ x 46 g/mol NO₂ = 51.5 g NO₂ will be produced

For part 2, use the Ideal Gas Law:

PV = nRT and solve for V (volume)

V = nRT/P = (1.12 moles)(0.0821 Latm/Kmol)(301K)/0.945 atm (change T to Kelvin and mbar to atm)

V = 29.3 L