Question

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

If 28.0 g of NH3 and 168.8 g of ClF3 are allowed to react, what masses of each product would be recovered assuming complete reaction?

Answer 1

There will be produced 58.14 grams of Cl2; 22.97 grams of N2 and 98.45 grams of HF

Step-by-step explanation:

Step 1: The balanced equation

2ClF3(g) + 2NH3(g) → N2(g) + Cl2(g) + 6HF(g)

Step 2: Data given

Mass of NH3 = 28.0 grams

Molar mass of NH3 = 17.03 grams

Mass of ClF3 = 168.8 grams

Molar mass ClF3 = 92.45 g/mol

Molar mass of N2 = 28.01 g/mol

Molar mass of Cl2 = 70.9 g/mol

Molar mass of HF = 20.01 g/mol

Step 3: Calculate moles

Moles of NH3 = mass NH3 / Molar mass NH3

Moles NH3 = 28.0 / 17.03 = 1.64 moles

Moles ClF3 = mass ClF3 / Molar mass ClF3

Moles ClF3 = 168.8 grams / 92.45 g/mol

Moles ClF3 = 1.83 moles

Step 4: Calculate limiting reactant

For 2 moles of NH3 we need 2 moles of ClF3 to produce 1 mole of Cl2, 1 mole of N2 and 6 moles of HF

NH3 is the limiting reactant. It will completely be consumed (1.64 moles).

ClF3 is in excess. There will be consumed 1.64 moles. There will remain 1.83 - 1.64 = 0.19 moles of ClF3

0.19 moles of ClF3 = 0.19 * 92.45 = 17.57 grams

Step 5: Calculate moles of Cl2, N2 and HF

For 2 moles of NH3 we need 2 moles of ClF3 to produce 1 mole of Cl2, 1 mole of N2 and 6 moles of HF

For 1.64 moles of NH3 consumed, we will produce:

1.64/ 2 = 0.82 moles of Cl2

1.64/2 = 0.82 moles of N2

1.64 * 3 = 4.92 moles of HF

Step 6: Calculate mass of products

Mass = moles * Molar mass

Mass of Cl2 = 0.82 moles * 70.9 g/mol = 58.138 grams

Mass of N2 = 0.82 moles * 28.01 g/mol = 22.97 grams

Mass of HF = 4.92 moles * 20.01 g/mol = 98.45 grams