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Answer:
(a)
Moles of ammonium chloride = 0.5 mole
Mass of ammonium chloride formed = 26.7455 g
(b)
Mole of = 0.125 mole
Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g
Mole of = 0.25 mole
Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g
Explanation:
(a)
For the first reaction:-
The mole ratio of the reactants = 1 : 1
0.5 moles of ammonia react with 0.5 moles of hydrochloric gas to give 0.5 moles of ammonium chloride
So, <u>Moles of ammonium chloride formed = 0.5 moles</u>
Molar mass of ammonium chloride = 53.491 g/mol
<u>Mass = Moles * Molar mass = 0.5 * 53.491 g = 26.7455 g</u>
(b)
For the first reaction:-
The mole ratio of the reactants = 1 : 4
It means
0.5 moles of methane react with 2.0 moles of sulfur to give 0.5 moles of Carbon disulfide and 1.0 moles of hydrogen sulfide gas.
But available moles of S = 0.5 moles
Limiting reagent is the one which is present in small amount. Thus, S is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
4 moles of S produces 1 mole of
Thus,
0.5 moles of S produces mole of
<u>Mole of = 0.125 mole</u>
Molar mass of = 76.139 g/mol
<u>Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g</u>
4 moles of S produces 2 moles of
Thus,
0.5 moles of S produces mole of
<u>Mole of = 0.25 mole</u>
Molar mass of = 34.1 g/mol
<u>Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g</u>
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