Question

1 ) Sulfur dioxide gas reacts with sodium hydroxide to form sodium sulfite and water. The unbalanced chemical equation for this reaction is given below:

SO₂(g) + NaOH(s) → Na₂SO₃(s) + H₂O(l)

Assuming that you start with 36.0 g of sulfur dioxide and 36.2 g of sodium hydroxide and assuming that the reaction goes to completion, determine the mass of each product formed.

g Na₂SO₃

g H₂O

2)

Ammonia gas reacts with sodium metal to form sodium amide (NaNH₂) and hydrogen gas. The unbalanced chemical equation for this reaction is given below:

NH₃(g) + Na(s) → NaNH₂(s) + H₂(g)

Assuming that you start with 32.4 g of ammonia gas and 10.4 g of sodium metal and assuming that the reaction goes to completion, determine the mass (in grams) of each product.

g NaNH₂

g H₂

3)

Consider the following unbalanced chemical equation.

H₂S(g) + O₂(g) → SO₂(g) + H₂O(g)

Determine the maximum number of moles of SO₂ produced from 11.0 moles of H₂S and 6.00 moles of O₂.

mol SO₂

Answer 1

1)

Molar mass of SO2 = 1*MM(S) + 2*MM(O)

= 1*32.07 + 2*16.0

= 64.07 g/mol

mass of SO2 = 36.0 g

we have below equation to be used:

number of mol of SO2,

n = mass of SO2/molar mass of SO2

=(36.0 g)/(64.07 g/mol)

= 0.5619 mol

Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass of NaOH = 36.2 g

we have below equation to be used:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(36.2 g)/(39.998 g/mol)

= 0.905 mol

we have the Balanced chemical equation as:

SO2 + 2 NaOH ---> Na2SO3 + H2O

1 mol of SO2 reacts with 2 mol of NaOH

for 0.5619 mol of SO2, 1.1238 mol of NaOH is required

But we have 0.905 mol of NaOH

so, NaOH is limiting reagent

we will use NaOH in further calculation

Molar mass of Na2SO3 = 2*MM(Na) + 1*MM(S) + 3*MM(O)

= 2*22.99 + 1*32.07 + 3*16.0

= 126.05 g/mol

From balanced chemical reaction, we see that

when 2 mol of NaOH reacts, 1 mol of Na2SO3 is formed

mol of Na2SO3 formed = (1/2)* moles of NaOH

= (1/2)*0.905

= 0.4525 mol

we have below equation to be used:

mass of Na2SO3 = number of mol * molar mass

= 0.4525*1.26*10^2

= 57.04 g

From balanced chemical reaction, we see that

when 2 mol of NaOH reacts, 1 mol of H2O is formed

mol of H2O formed = (1/2)* moles of NaOH

= (1/2)*0.905

= 0.4525 mol

we have below equation to be used:

mass of H2O = number of mol * molar mass

= 0.4525*18

= 8.145 g

57.0 g Na2SO3

8.15 g H2O

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