Solvay process is used to manufacture sodium carbonate. During this process ammonia isrecovered by the following reaction. (3+3)2NH4Cl + Ca(OH)2 CaCl2 + 2H2O +2NH3When 100 g of ammonium chloride and 150 g calcium hydroxide are used theni. Calculate the mass in kg of ammonia produce during chemical reaction.ii. Calculate the excess mass in gram of one of the reactants left unreacted.(At. Mass N=14 H=1 Cl= 35.5 Ca=40)

First, we need to balance the chemical equation. The balanced equation is:

2NH4Cl + Ca(OH)2 → CaCl2 + 2H2O + 2NH3

i. To calculate the mass of ammonia produced, we first need to find the limiting reactant.

The molar mass of NH4Cl is 53.5 g/mol (142 + 14 + 35.5) and for Ca(OH)2 it is 74 g/mol (40 + 162 + 12).

So, the number of moles of NH4Cl = 100 g / 53.5 g/mol = 1.87 mol And the number of moles of Ca(OH)2 = 150 g / 74 g/mol = 2.03 mol

From the balanced equation, we can see that 2 moles of NH4Cl react with 1 mole of Ca(OH)2. So, NH4Cl is the limiting reactant.

The balanced equation also tells us that 2 moles of NH4Cl produce 2 moles of NH3. So, 1.87 moles of NH4Cl will produce 1.87 moles of NH3.

The molar mass of NH3 is 17 g/mol (14 + 1*3). So, the mass of NH3 produced = 1.87 mol * 17 g/mol = 31.79 g or 0.03179 kg.

ii. To find the excess reactant, we subtract the amount of Ca(OH)2 used up from the total amount.

From the balanced equation, we know that 2 moles of NH4Cl react with 1 mole of Ca(OH)2. So, 1.87 moles of NH4Cl will react with 1.87/2 = 0.935 moles of Ca(OH)2.

The number of moles of Ca(OH)2 left = total moles - moles used = 2.03 mol - 0.935 mol = 1.095 mol

So, the mass of Ca(OH)2 left = 1.095 mol * 74 g/mol = 81.03 g.