Answer to Question #233849 in Chemistry for thegifinator

Question

205 gram hydrogen sulfide

reacted with 75.0 g oxygen to

produce sulfur and water.

Find:

a) Gram of sulfur formed

b) gram of excess reactant

used up

c) gram of excess reactant

that remained unreacted

2. 295 gram Rubidium Oxide

reacted with 202 gram calcium

cyanide to produce Rubidium

cyanide and Calcium Oxide.

Find

a)gram of Calcium oxide

b) gram of excess reactant

used up

c) gram of excess reactant that

remained unreacted

Answer 1

Solution (1):

Balanced cheical equation:

2H₂S + O₂ → 2S + 2H₂O

Determine moles of each reactant:

(205 g H₂S) × (1 mol H₂S / 34.1 g H₂S) = 6.012 mol H₂S

(75.0 g O₂) × (1 mol O₂ / 31.998 g O₂) = 2.344 mol O₂

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it.

Let's choose O₂:

n(H₂S) = 2 × n(O₂) = (2 × 2.344) = 4.688 mol

The calculation above means that we need 4.688 mol of H₂S to completely react with O₂.

We have 6.012 mol of H₂S and therefore more than enough hydrogen sulfide.

Thus hydrogen sulfide (H₂S) is in excess and oxygen (O₂) must be the limiting reagent.

Hence, O₂ will run out first.

Therefore,

a)

n(S) = 2 × n(O₂) = (2 × 2.344) = 4.688 mol

(4.688 mol S) × (32.065 g S / 1 mol S) = 150.32 g S = 150 g S (formed)

b)

n(H₂S) = 4.688 mol

(4.688 mol H₂S) × (34.1 g H₂S / 1 mol H₂S) = 159.86 g H₂S = 160 g H₂S (used up)

c)

205 g H₂S - 160 g H₂S = 45.0 g H₂S (remained unreacted)

Answer (1):

a) 150 grams of sulfur (S) formed.

b) 160 grams of excess reactant (H₂S) used up.

c) 45.0 grams of excess reactant (H₂S) remained unreacted

Solution(2):

Balanced cheical equation:

Rb₂O + Ca(CN)₂ → 2RbCN + CaO

Determine moles of each reactant:

(295 g Rb₂O) × (1 mol Rb₂O / 186.94 g Rb₂O) = 1.578 mol Rb₂O

(202 g Ca(CN)₂) × (1 mol Ca(CN)₂ / 92.11 g Ca(CN)₂) = 2.193 mol Ca(CN)₂

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it.

Let's choose Rb₂O:

n(Ca(CN)₂) = n(Rb₂O) = 1.578 mol

The calculation above means that we need 1.578 mol of Ca(CN)₂ to completely react with Rb₂O.

We have 2.193 mol of Ca(CN)₂ and therefore more than enough calcium cyanide.

Thus calcium cyanide (Ca(CN)₂) is in excess and rubidium oxide (Rb₂O) must be the limiting reagent.

Hence, Rb₂O will run out first.

Therefore,

a)

n(CaO) = n(Rb₂O) = 1.578 mol

(1.578 mol CaO) × (56.0774 g CaO / 1 mol CaO) = 88.49 g CaO = 88.5 g CaO (formed)

b)

n(Ca(CN)₂) = 1.578 mol

(1.578 mol Ca(CN)₂) × (92.11 g Ca(CN)₂ / 1 mol Ca(CN)₂) = 145.35 g Ca(CN)₂ = 145 g Ca(CN)₂ (used up)

c)

202 g Ca(CN)₂ - 145 g Ca(CN)₂ = 57.0 g Ca(CN)₂ (remained unreacted)

Answer (2):

a) 88.5 grams of calcium oxide (CaO) formed.

b) 145 grams of excess reactant (Ca(CN)₂) used up.

c) 57.0 grams of excess reactant (Ca(CN)₂) remained unreacted