Question

For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be produced by the complete conversion of 0.50 mole of the reactant indicated in boldface. State clearly the mole ratio used for each conversion.
a. NH3(g)+HCL(g)->NH4CL(s)
b. CH4(g)+4S(s)->CS2(L)+2H2S(g)

Answer 1

Answer:

(a)

Moles of ammonium chloride = 0.5 mole

Mass of ammonium chloride formed = 26.7455 g

(b)

Mole of $CS_2$ = 0.125 mole

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

Mole of $H_2S$ = 0.25 mole

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

Explanation:

(a)

For the first reaction:-

$NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}$

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, Moles of ammonium chloride formed = 0.5 moles

Molar mass of ammonium chloride = 53.491 g/mol

Mass = Moles * Molar mass = 0.5 * 53.491 g = 26.7455 g

(b)

For the first reaction:-

$CH_4_{(g)}+4S_{(s)}\rightarrow CS_2_{(l)}+2H_2S_{(g)}$

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 $CS_2$

Thus,

0.5 moles of S produces $\frac{1}{4}\times 0.5$ mole of $CS_2$

Mole of $CS_2$ = 0.125 mole

Molar mass of $CS_2$ = 76.139 g/mol

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

4 moles of S produces 2 moles of $H_2S$

Thus,

0.5 moles of S produces $\frac{2}{4}\times 0.5$ mole of $H_2S$

Mole of $H_2S$ = 0.25 mole

Molar mass of $H_2S$ = 34.1 g/mol

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g