Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.397 moles of nitrogen monoxide are mixed with 0.307 moles of oxygen gas.

Question

What is the formula for the limiting reagent?

What is the maximum amount of nitrogen dioxide that can be produced?

Answer 1

2NO + O₂ = 2NO₂

Here, 2 moles of NO react with 1 mole of O₂. In order to react 0.307 moles of oxygen gas, we need (0.307 * 2) or 0.614 moles of NO(which we do not have). On the other hand, in order to consume 0.397 moles NO, we need (0.397/2) or 0.1985 moles of O₂. So, NO is the limiting reagent as it is consumed before O₂.

1 mole of NO produces 1 mole of NO₂. 0.397 mole NO produces 0.397 mole of NO₂.

molar mass of NO₂ = 14 + (16 * 2) = 46 g/mol

mass of NO₂ = number of moles * molar mass = 0.397 * 46 = 18.262 g