Question

According to the following reaction, how many grams of nitrogen monoxide will be formed upon the complete reaction of 31.0 grams of oxygen gas with excess ammonia? ammonia (g) + oxygen (g) nitrogen monoxide (g) + water (g)

Answer 1

23.227 grams of NO is formed upon the complete reaction of 31.0 grams of oxygen gas with excess ammonia.

Explanation:

The balanced chemical equation is:

4NH3 + 5O2 ⇒ 4NO + 6H20

Number of moles will be calculated.

31 grams of oxygen is given

atomic weight of oxygen gas i.e O2 = 32gm/mole

The molar mass of oxygen gas is 32 grams/mole which is equal to one mole of the molecule.

Number of moles (n) = $\frac{mass}{atomic mass of one mole of the substance}$

                             n=    $\frac{31}{32}$

                              n = 0.968 moles of oxygen are given for the reaction to occur.

From the balanced chemical equation it can be seen

5 moles of oxygen yielded 4 moles of NO

So, 0.968 moles of oxygen yield x moles of NO

$\frac{4}{5}$ = $\frac{x}{0.968 }$

4 × 0.968 =5x

x = 0.774 moles of NO will be formed.

To calculate the mass of NO

mass= atomic mass x number of moles

atomic mass of NO is 30.01 grams/mole

putting the values in the formula:

mass = 30.01 x 0.774

          = 23.227 grams of NO is formed.