5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.
<u>Explanation</u>:
- Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:
Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃
- From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of H₂SO₄.
- Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.
- To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is = 5.451 X 10³ kg.
Answer:
Explanation:
From the question, we have been asked to find the molarity of FeCl2 having a volume of 450 mL,
We have been provided with 225 g which is proportional to 1.8 moles.
We know that molarity of any solution should be in mol/L.
1 mole contained in 1 L means it has a molarity of 1 mol/L
Let's convert 450 mL to Litres which is,
= 0.450 L
Thus,
1 mole is contained in 1L
x moles are contained in 0.450 L
Hence,
x mole/molarity = {1 mole x 1 L}/{0.450 L}
= 4 mol/L
Therefore 4 mol/L is the molar concentration.
The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL
<h3>Balanced equation </h3>
2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HCl (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 1
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
- Volume of acid, HCl (Va) = 36 mL
- Molarity of acid, HCl (Ma) = 1.63 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(1.63 × 36) / (1.48 × Vb) = 2
58.68 / (1.48 × Vb) = 2
Cross multiply
2 × 1.48 × Vb = 58.68
2.96 × Vb = 58.68
Divide both side by 2.96
Vb = 58.68 / 2.96
Vb = 19.8 mL
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Answer:
Away from the central sulfur atom.
Explanation:
