Solvay process involves these reactions...questions follow...?

Question

1. The Solvay process for the manufacture of sodium carbonate involves the following reactions:
NH3(g) + H2O(l) + CO2(g)-(arrow) NH4HCO3(aq)
NH4HCO3(aq) + NaCl(aq)-(arrow) NH4Cl(aq) + NaHCO3(s)
2NaHCO3(s) -(arrow) Na2CO3(s) + H2O(g) + CO2(g)

Calculate the masses of ammonia and sodium chloride required to produce 150.0kg of pure anhydrous sodium carbonate.

2. A 2.750g sample of zinc carbonate was treated with a solution containing 0.0500 moles of hydrochloric acid.
ZnCO3(s) + 2HCl(aq) -(arrow) ZnCl2(aq) + H2O(l)+ CO2(g)
Find the
a) limiting reactant
b) mass of carbon dioxide produced
c) mass of zinc chloride that could be recovered
d) mass of remaining reactant

Thanks for helping. Please show working and explain it to me if possible. Thanks so much!

Answer 1

1) In a way, part of this is a trick question because in the Solvay process, the Ammonia ia recycled in an equation that you left out of the sequence:

2 NH4Cl + CaO --> 2 NH3 + CaCl2 + H2O

Ammonia is considered a catalyst, so it is not used up in the process. However, assuming that no Ammonia is recycled, the production of 150.0 Kg of anhydrous Sodium carbonate means that:

150.0 Kg / 106.0 g/mol = 150.0 Kg / 0.106 Kg/mol
= 1415 moles of Na2CO3

This would require twice as many moles of Ammonia and NaCl to make, or 1415 moles * 2 = 2830 moles.

For the Ammonia, 2830 moles * 17.0304 g/mol = 48,196 g or 48.2 Kg.

For the Sodium chloride, 2830 moles * 58.442 g/mol = 165,390 g or 165.4 Kg.

2) For: ZnCO3 (s) + 2 HCl (aq) --> ZnCl2 (aq) + H2O (l) + CO2 (g)

One mole of the Carbonate reacts with 2 moles of the Acid.

a) 2.750 g / MW ZnCO3 = moles ZnCO3
compare this with 0.25 moles HCl
Figure the actual moles of ZnCO3 used.
b) moles of ZnCO3 = moles of CO2
Moles of CO2 * MW CO2 = mass of CO2
c) moles of ZnCO3 = moles of ZnCl2
Moles of ZnCl2 * MW ZnCl2 = mass of ZnCl2
d) from limiting reactant, figure one in excesss and calculate starting amount minus ending amount.