Question

How many grams of `H_2` are needed to produce 11.27 g of `NH_3`?

Answer 1

We need (i) the stoichiometric equation, and (ii) the equivalent mass of dihydrogen.

Explanation:

`1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)`

`11.27` `g` of ammonia represents `(11.27*g)/(17.03*g*mol^-1)` `=` `??` `mol`.

Whatever this molar quantity is, it is clear from the stoichiometry of the reaction that 3/2 equiv of dihydrogen gas were required. How much dinitrogen gas was required?

Answer 2

2.001 g hydrogen gas is needed to produce 11.27 g ammonia if nitrogen gas is present in excess.

Explanation:

Start with a balanced equation.

`"3H"_2+"N"_2` `rarr` `"2NH"_3"`

Since the question concerns the mass of hydrogen gas needed, the assumption is that nitrogen gas is in excess.

The molar masses of `"H"_2"` and `"NH"_3"` are needed.

`"H"_2":` `"2.01588 g/mol"`
http://pubchem.ncbi.nlm.nih.gov/compound/783

`"NH"_3"` `"17.03052 g/mol"`
http://pubchem.ncbi.nlm.nih.gov/compound/222#section=Top

We will be making the following conversions: mass `"NH"_3"` `rarr` `"mol NH"_3"` `rarr` `"mol H"_2"` `rarr` `"mass H"_2"`

1. Determine moles of `"NH"_3"` by dividing the given mass by its molar mass.

2. Determine moles of `"H"_2"` by multiplying the mole ratio from the balanced equation between `"H"_2"` and `"NH"_3"` with moles `"H"_2"` in the numerator. `(3"mol H"_2)/(2"mol NH"_3)`

3. Determine mass of `"H"_2"` by multiplying moles `"H"_2"` by its molar mass.

`11.27"g NH"_3xx(1"mol NH"_3)/(17.03052"g NH"_3)xx(3"mol H"_2)/(2"mol NH"_3)xx(2.01588"g H"_2)/(1"mol H"_2)="2.001 g H"_2"` (rounded to four significant figures)

2.001 g hydrogen gas is needed to produce 11.27 g ammonia if nitrogen gas is present in excess.