Question

For the following reaction, 10.4 grams of nitrogen monoxide are allowed to react with 8.80 grams of oxygen gas . nitrogen monoxide(g) + oxygen(g) nitrogen dioxide(g) What is the maximum mass of nitrogen dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

Balanced equation is

2NO + O2 -----> 2 NO2

number of moles of NO = 10.4g / 30.01 g/mol = 0.347 mole

Number of moles of O2 = 8.80g / 32.0 g/mol = 0.275 mole

from tha balanced equation we can say that

2 mole of NO requires 1 mole of O2 so

0.347 mole of NO will require 0.1735 mole of O2

but we have 0.275 mole of O2 which is in excess so NO is the limiting reactant

The formula of limiting reactant = NO

from the balanced equation we can say that

2 mole of NO produces 2 mole of NO2 so

0.347 mole of NO will produce 0.347 mole of NO2

1 mole of NO2 = 46.0055 g

0.347 moel of NO2 = 16.00 g

Therefore, the mass of NO2 produced = 16.00g

The number of moles of excess reactant = 0.275 - 0.1735 = 0.1015 mole

1 mole of O2 = 32.0 g

0.1015 mole of O2 = 3.248 g

Therefore, the mass of excess reactant = 3.248 g