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2 years ago

1 how many moles of sodium bicarbonate are needed to neutralize 0.9ml of sulphuric acid at stp

Chemistry

1 answer:

Natali5045456 [20]2 years ago

3 0

Answer:

8.0356 * 10^-5 moles of NaHCO3

Explanation:

Sulphuric acid = H2SO4

Sodium bicarbonate = NaHCO3

The reaction between both compounds is given by;

2NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)

In the reactin above;

2 mol of NaHCO3 neutralizes 1 mol of H2SO4

At stp, 1 mol occupies 22.4 L;

1 mol = 22.4 L = 22400 mL

x mol = 0.9 mL

x = 0.9 / 22400 = 4.0178 * 10^-5 moles of H2SO4

Since 2 mol = 1 mol from the equation;

x mol = 4.0178 * 10^-5

x mol = 2 * 4.0178 * 10^-5

x = 8.0356 * 10^-5 moles of NaHCO3

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Answer:

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Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

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2 years ago

1 how many moles of sodium bicarbonate are needed to neutralize 0.9ml of sulphuric acid at stp​

1 how many moles of sodium bicarbonate are needed to neutralize 0.9ml of sulphuric acid at stp