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Luden [163]
3 years ago
9

30 pts! Sodium and water react according to the following equation. If 31.5g of sodium are added to excess water, how many liter

s of hydrogen gas are formed at STP? Show all work for credit. 2Na+2H2O--> 2NaOH+H2
Chemistry
1 answer:
mihalych1998 [28]3 years ago
3 0
First, calculate the number of moles of sodium present with the given mass,

              31.5 g of sodium x (1 mol sodium/ 23 g sodium) = 1.37 mol sodium

It is given in the equation that for every 2mols of sodium, one mol of H2 is produced.

           mols of H2 = (1.37 mols sodium)(1 mol H2/ 2 mols sodium)
                  mols of H2 = 0.685 mols H2

Then, at STP, 1 mol of gas = 22.4 L.
                
                 volume of H2 = (0.685 mols H2)(22.4 L / 1 mol)
                     volume of H2 = 15.34 L

Answer: 15.34 L
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2 years ago
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A student was given a 50.0 mL of 0.10 M solution of an unknown diprotic acid, H2A, which was titrated with 0.10 M NaOH. After a
stich3 [128]

Answer:

Ka1 = 2.00x10⁻⁷, Ka2 = 5.00x10⁻¹⁰

Explanation:

A diprotic acid is a substance that can release 2 H⁺ when in aqueous solution. Because it is a weak acid, the ionization will be reversible. So, the acid has two equilibrium reactions, each one with its equilibrium constant:

H₂A ⇄ H⁺ + HA⁻   Ka1 = ([HA⁻]*[H⁺]/[H₂A])

HA⁻ ⇄ H⁺ + A⁻     Ka2 = ([A⁻]*[H⁺]/[HA⁻])

First, the number of moles of H₂A was:

n = 0.10 mol/L *0.05L = 0.005 mol

And then was added NaOH:

n = 0.1 mol/L * 0.025 L = 0.0025 mol

So, all the NaOH will reacts, then, the number of moles will be:

H₂A = 0.005 - 0.0025 = 0.0025 mol

HA⁻ = 0.0025 mol (the stoichiometry is 1:1:1)

The concentrations of H₂A and HA⁻ will be the same, so Ka1 = [H⁺], and

pH = -log[H⁺]

6.70 = -log[H⁺]

[H⁺] = 10^{-6.70}

[H⁺] = 2.00x10⁻⁷ M

Ka1 = 2.00x10⁻⁷

After the addition of 50.0 mL of NaOH the second equilibrium must dominate the reaction. For the expression of Ka1:

Ka1 = ([HA⁻]*[H⁺]/[H₂A])

[HA⁻]*[H⁺] = Ka1*[H₂A]

[HA⁻] = (Ka1*[H₂A])/[H⁺]

Ka2 = ([A⁻]*[H⁺]/[HA⁻])

Ka2 = ([A⁻]*[H⁺]²)/(Ka1*[H₂A])

By the stoichiometry, [H₂A] = [A⁻], so:

Ka2 = [H⁺]²/Ka1

pH = -log[H⁺]

8.00 = -log[H⁺]

[H+] = 10⁻⁸

Ka2 = (10⁻⁸)²/(2.00x10⁻⁷)

Ka2 = 5.00x10⁻¹⁰

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