Question

Given the following balanced reaction between liquid ammonia and oxygen gas to produce nitrous oxide gas and water, how many grams of water, H2O, are produced from 317 grams of ammonia and excess oxygen? (To find the molar mass in the problem use the periodic table and round the mass to the hundreds place for calculation).



(A) 224 g
(B) 335 g
(C) 503 g

Answer 1

Answer:

(C) 503 g

Explanation:

Balanced reaction:

2NH₃ + 2O₂ → N₂O + 3H₂O

Stoichiometry:

317 g NH₃ × (1 mol NH₃ / 17.03 g NH₃) = 18.61 mol NH₃

18.61 mol NH₃ × (3 mol H₂O / 2 mol NH₃) = 27.92 mol H₂O

27.92 mol H₂O × (18.02 g H₂O / mol H₂O) = 503 g H₂O

Answer 2

Answer:

(C) 503 g of water, H2O, are produced from 317 grams of ammonia and excess oxygen

Explanation:

$2 NH_3+2 O_2>N_2 O+3 H_2 O$

Molar mass of

$H_2 O=(2\times1.008)+15.999=18.02 g/mol$

Molar mass of

$NH_3=14.01+(3\times1.008)=17.03 g/mol$

The conversions are

Step 1:

Mass $NH_3$ to moles $NH_3$ by dividing with molar mass $NH_3$

Step 2:

Moles $NH_3$ to moles $H_2 O$ by using mole ratio of $NH_3:H_2 O$ i.e., 2 : 3

Step 3:

Moles $H_2 O$ to mass $H_2 O$ by multiplying with molar mass $H_2 O$

$317gNH_3  \times \frac {(1mol NH_3)}{(17.03gNH_3 )} \times \frac {(3mol H_2 O)}{(2mol NH_3 )} \times \frac {(18.02g H_2 O)}{(1mol H_2 O)}$

$= 503g H_2 O$ is formed.

(Answer)

Please note :

To convert moles to mass, we multiply by molar mass  

To convert mass to moles, we divide by molar mass

Molar mass is the mass of 1 mole of the substance

For example molar mass of $CO_2$ is

$(12.0+2\times16)=44 g/mol$

(we just add the atomic mass of the atoms to get the molar mass of the substance).