<h3>
Answer:</h3>
266.325 g
<h3>
Explanation:</h3>
We are given the balanced equation;
2NaOH + H₂SO₄ → H₂O + Na₂SO₄
We are required to determine the mass of Na₂SO₄ that will be formed.
<h3>Step 1: Determine the number of moles of NaOH</h3>
Moles = Mass ÷ molar mass
Molar mass of NaOH is 40.0 g/mol
Therefore;
Moles of NaOH = 150 g ÷ 40 g/mol
= 3.75 moles
<h3>Step 2: Determine the number of moles of sodium sulfate formed</h3>
- From the equation 2 moles of NaOH reacts with sulfuric acid to form 1 mole of sodium sulfate.
- Therefore; mole ratio of NaOH : Na₂SO₄ is 2 : 1
Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2
= 3.75 moles ÷ 2
= 1.875 moles
<h3>Step 3: Determine the mass of Na₂SO₄ produced.</h3>
we know that;
Mass = Moles × Molar mass
Molar mass of Na₂SO₄ is 142.04 g/mol
Therefore;
Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol
= 266.325 g
Thus, the mass of sodium sulfate formed 266.325 g
Answer:
120.37 g
Explanation:
Mg = 24.31 g
S = 32.06 g
4 O = 16 X 4 = 64 g
Therefore, MgSO4 is (24.31 + 32.06 + 64) g = 120.37 g
Answer:
the solubility of CaCO3 is 0.015g/l 25 °C
is favored at equilibrium
Explanation:
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?
solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)
CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)
in partial dissociation , we can say
2.25x 10^-8=
let Ca^2+=CO3^-2=S
2.25x10^-8=S*S
S^2=2.25x10^-8
S=0.00015mol/L
Converting that to g/l
the relative molecular mass of CaCO3=100g/mol
0.00015*100g/mol
0.015g/l
the solubility of CaCO3 is 0.015g/l @room temperature
is favored at equilibrium
Answer:
unknown fact I can't message anyone
