Question

How many grams of H2 are needed to produce 10.42 g of NH3

Answer 1

N2 + 3 H2 >> 2 NH3 moles NH3 = 11.50 g /17.0307 g/mol=0.6753 the ratio between H2 and NH3 is 3 : 2 moles H2 needed = 0.6753 x 3/2 =1.013 mass H2 = 1.013 mol x 2.106 g/mol=2.042 g

Answer 2

Answer:

1.84g of H₂ are needed to produce 10.42g of NH₃

Explanation:

In the production of ammonia, NH₃, the following reaction occurs

N₂ + 3H₂ ⇄ 2NH₃

From, the above, 3 moles of hydrogen gas (H₂) is required to produce 2 moles of ammonia (NH₃).

The atomic mass of Hydrogen is 1 g/mol, 3 moles of hydrogen gas are present in the reactant, hence the mass of hydrogen gas in the equation is;

3 × 2 × 1 = 6g

The atomic mass of Nitrogen is 14g/mol, that of hydrogen is 1g/mol and 2 moles of ammonia were produced. Hence the mass of ammonia in the equation is;

(2 × 14) + (2 × 3 × 1) = 34g

From the above, it can be deduced that 6g of H₂ is required to produce 34g of NH₃, hence "x" grams of H₂ are needed to produce 10.42g of NH₃

To find x (which is the unknown)

6g ⇒ 34g

x ⇒ 10.42g

x = $\frac{10.42 * 6}{34}$

x = 1.84g

1.84g of H₂ are needed to produce 10.42g of NH₃