Question

Part b what is the maximum mass, in grams, of nh3 that can be produced by the reaction of 1.0 g of n2 and 1.0 g of h2 using the reaction below? n2+h2?nh3 (not balanced)

Answer 1

Answer: The maximum mass of ammonia produced will be 1.224 grams

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For hydrogen gas:

Given mass of hydrogen gas = 1.0 g

Molar mass of hydrogen gas = 2.0 g/mol

Putting values in equation 1, we get:

`\text{Moles of hydrogen gas}=(1.0g)/(2.0g/mol)=0.5mol`

• For nitrogen gas:

Given mass of nitrogen gas = 1.0 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

`\text{Moles of nitrogen gas}=(1.0g)/(28g/mol)=0.036mol`

The chemical equation for the formation of ammonia follows:

`3H_2+N_2\rightarrow 2NH_3`

By Stoichiometry of the reaction:

1 mole of nitrogen gas reacts with 3 moles of hydrogen gas

So, 0.036 moles of nitrogen gas will react with =
`(3)/(1)* 0.036=0.108mol` of hydrogen gas

As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of nitrogen gas produces 2 moles of ammonia.

So, 0.036 moles of nitrogen gas will produce
`(2)/(1)* 0.036=0.072mol` of ammonia

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 0.072 moles

Putting values in equation 1, we get:

`0.072mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(0.072mol* 17g/mol)=1.224g`

Hence, the maximum mass of ammonia produced will be 1.224 grams

Answer 2

3H₂ + N₂ = 2NH₃

n(H₂)=m(H₂)/M(H₂)
n(H₂)=1.0/2.0=0.5 mol

n(N₂)=m(N₂)/M(N₂)
n(N₂)=1.0/28.0=0.0357 mol

H₂:N₂ = 3:1
0.5:0.0357 = 3:0.2142 nitrogen is a limiting reagent

n(NH₃)=2n(N₂)
m(NH₃)=2n(N₂)M(NH₃)
m(NH₃)=2*0.0357*17.0=1.214 g