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postnew [5]
3 years ago
14

Suppose you mix one mole of sulfuric acid (H2SO4) with 1 mole of sodium hydroxide(NaOH).

Chemistry
1 answer:
olasank [31]3 years ago
7 0

Answer & Explanation:

  • The neutralization of H₂SO₄ with NaOH is occurred according to the balanced equation:

<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O,</em>

It is clear that every 1.0 mol of H₂SO₄ needs 2 mol of NaOH to be neutralized completely.

<em>So, when you mix one mole of sulfuric acid with 1 mole of sodium hydroxide, there will be an excess of sulfuric acid.</em>

<em>Thus, the pH of the solution remain below 7.</em>

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How many grams of Sg is required to produce 83.10 g SF6? S: +24F--&gt;8SF
ozzi

Answer : The mass of S_8 required is 18.238 grams.

Explanation : Given,

Mass of SF_6 = 83.10 g

Molar mass of SF_6 = 146 g/mole

Molar mass of S_8 = 256.52 g/mole

The balanced chemical reaction is,

S_8+24F_2\rightarrow 8SF_6

First we have to determine the moles of SF_6.

\text{Moles of }SF_6=\frac{\text{Mass of }SF_6}{\text{Molar mass of }SF_6}=\frac{83.10g}{146g/mole}=0.569moles

Now we have to determine the moles of S_8.

From the balanced chemical reaction we conclude that,

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So, 0.569 moles of SF_6 produced from \frac{0.569}{8}=0.0711 mole of S_8

Now we have to determine the mass of S_8.

\text{Mass of }S_8=\text{Moles of }S_8\times \text{Molar mass of }S_8

\text{Mass of }S_8=(0.0711mole)\times (256.52g/mole)=18.238g

Therefore, the mass of S_8 required is 18.238 grams.

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