Answer:
A.
Explanation:
Adjust = Adapt if that makes sense.
The volume of H₂ : = 15.2208 L
<h3>Further explanation</h3>
Given
Reaction
2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)
34.0g of As
Required
The volume of H₂ at STP
Solution
mol As (Ar = 75 g/mol) :
= mass : Ar
= 34 g : 75 g/mol
= 0.453 mol
From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :
=3/2 x mol As
=3/2 x 0.453
= 0.6795
At STP, 1 mol = 22.4 L, so :
= 0.6795 x 22.4 L
= 15.2208 L
Answer:
2.2 moles of Fe will be produced
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 3.3 moles
Number of moles of iron oxide = 1.5 moles
Step 2: The balanced equation
3H2 + Fe2O3 → 2Fe + 3H2O
Step 3: Calculate the limiting reactant
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
Hydrogen gas is the limiting reactant. It will completely be consumed (3.3 moles). Fe2O3 is in excess. There will react 3.3 / 3 = 1.1 moles
There will remain 1.5 - 1.1 = 0.4 moles Fe2O3
Step 4: Calculate moles Fe
For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O
For 3.3 moles H2 we'll have 2/3 * 3.3 = 2.2 moles Fe
2.2 moles of Fe will be produced
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
Answer:

Explanation:
You must convert 30 % (m/v) to a molar concentration.
Assume 1 L of solution.
1. Mass of NaOH

2. Moles of NaOH

3. Molar concentration of NaOH

4. Volume of NaOH
Now that you know the concentration, you can use the dilution formula .

to calculate the volume of stock solution.
Data:
c₁ = 7.50 mol·L⁻¹; V₁ = ?
c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL
Calculations:
(a) Convert millilitres to litres

(b) Calculate the volume of dilute solution

