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patriot [66]
3 years ago
14

If 100. mL of 0.400 M Na2SO4 is added to 200. mL of 0.600 M NaCl, what is the concentration of Na+ ions in the final solution? A

ssume that the volumes are additive.
Chemistry
1 answer:
Dmitrij [34]3 years ago
4 0
Na₂SO₄ → 2Na⁺ + SO₄²⁻

c₁(Na⁺)=2c(Na₂SO₄)

n₁(Na⁺)=c₁(Na⁺)v₁=2c(Na₂SO₄)v₁

NaCl → Na⁺ + Cl⁻

c₂(Na⁺)=c(NaCl)

n₂(Na⁺)=c₂(Na⁺)v₂=c(NaCl)v₂

n(Na⁺)=n₁(Na⁺)+n₂(Na⁺)=2c(Na₂SO₄)v₁+c(NaCl)v₂

v=v₁+v₂

c(Na⁺)=n(Na⁺)/v={2c(Na₂SO₄)v₁+c(NaCl)v₂}/(v₁+v₂)

c(Na⁺)={2*0.400*0.1+0.600*0.2}/(0.1+0.2)=0.667 mol/L

c(Na⁺)=0.667M
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Answer:

We need 226 grams of FeS

Explanation:

Step 1: Data given

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Step 2: The balanced equation

FeS + 2 HCl → H2S + FeCl2

Step 3: Calculate moles FeCl2

Moles FeCl2 = 326 grams / 126.75 grams

Moles FeCl2 = 2.57 moles

Step 4: Calculate moles FeS needed

For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl

For 2.57 moles FeCl2 we need 2.57 moles FeS

Step 5: Calculate mass FeS

Mass FeS = 2.57 moles * 87.92 g/mol

Mass FeS = 226 grams FeS

We need 226 grams of FeS

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