Question

Help me with this please i need chem help lol

Answer 1

`\boxed{\text{1. 0.47 mol; 2. 0.046 mol; 3 (a) 11.6 mol, 15.4 mol, (b) 2.31 mol, 3.08 mol, 0.770 mol}}`

Step-by-step explanation:

1. Moles of Na

(a) Balanced equation

2Na + 2H₂O ⟶ 2NaOH + H₂

(b) Calculation

You want to convert moles of H₂ to moles of Na

The molar ratio is 2 mol Na:1 mol H₂

Moles of Na = 4.0 mol H₂ × (2 mol Na/1 mol H₂) = 8.0 mol Na

You need
`\boxed{ \text{8.0 mol of Na}}` to form 4.0 mol of H₂.

2. Moles of LiCl

(a) Balanced equation

2LiBr + Cl₂⟶ 2LiCl + Br₂

(b) Calculation

You want to convert moles of LiBr to moles of LiCl

The molar ratio is 2 mol LiBr:2 mol LiCl

Moles of LiCl = 0.046 mol LiBr × (2 mol LiCl/2 mol LiBr) = 0.046 mol LiCl

The reaction will produce
`\boxed{ \text{0.046 mol of LiCl}}`.

3. Combustion of propane

C₃H₈ +5O₂ ⟶ 3CO₂ +4H₂O

(a) Moles of CO₂ and H₂O

Moles of CO₂ = 0.647 mol O₂ × (3 mol CO₂/1 mol C₃H₈) = 11.6 mol CO₂

Moles of H₂O = 3.85 mol O₂ × (4 mol CO₂/1 mol C₃H₈) = 15.4 mol H₂O

The reaction produces
`\boxed{ \text{11.6 mol of CO}_(2)}` and
`\boxed{ \text{15.4 mol of H}_(2)\text{O}}`.

(b) Moles from O₂

Moles of CO₂ = 3.85 mol O₂ × (3 mol CO₂/5 mol O₂) = 2.31 mol CO₂

Moles of H₂O = 3.85 mol O₂ × (4 mol CO₂/5 mol O₂) = 3.08 mol H₂O

Moles of C₃H₈ = 3.85 mol O₂ × (1 mol C₃H₈/5 mol O₂) = 0.770 mol C₃H₈

The reaction produces
`\boxed{ \text{2.31 mol}}` of CO₂,
`\boxed{ \text{3.08 mol}}` of H₂O, and consumes
`\boxed{ \text{0.770 mol}}` of C₃H₈.