Question

a. How many grams of O2 are needed to react with 13.6 g
of NH3?
b. How many grams of N2 can be produced when 6.50 g
of O2 reacts?
c. How many grams of H20 are formed from the reaction of
34.0 g of NH3?

Answer 1

a.Answer:

Mass = 32 g

Explanation:

Given data:

Mass of O₂ needed = ?

Mass of NH₃ react = 13.6 g

Solution:

Chemical equation:

4NH₃ + 5O₂   →  4NO + 6H₂O

Number of moles of ammonia:

Number of moles = mass/ molar mass

Number of moles = 13.6 g/ 17 g/mol

Number of moles = 0.8 mol

Now we will compare the moles of ammonia and oxygen from balance chemical equation.

                        NH₃          :            O₂

                          4            :               5

                         0.8          :         5/4×0.8 = 1

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 1 mol × 32 g/mol

Mass = 32 g

b.Answer:

3.79 g of N2.

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

4NH3 + 3O2 → 2N2 + 6H2O

Next, we shall determine the mass of O2 that reacted and the mass of N2 produced from the balanced equation.

This is illustrated below:

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Summary:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Finally, we shall determine the mass of N2 produced by reacting 6.50 g of O2.

This can be obtained as follow:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.50 g of O2 will react to produce = (6.50 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were obtained from the reaction..

c.That gives you 18.16 g of water.