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qaws [65]
3 years ago
12

The reaction of sodium peroxide and water produces sodium hydroxide and oxygen gas. The following balanced chemical equation rep

resents the reaction.
2 Na^2O^2(s) + 2 H2O(l) → 4 NaOH(s) + O^2(g)

If 15.7 moles of sodium hydroxide are produced, how many moles of O2 will be made?
Chemistry
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

3.925 mol.

Explanation:

  • From the balanced equation:

<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g) ,</em>

It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂ .

<em>Using cross multiplication:</em>

4 moles of NaOH produced with → 1 mole of O₂ .

15.7 moles of NaOH produced with → ??? mole of O₂ .

<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) =  <em>3.925 mol.</em>

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a. blue, b. red, c. red, d. basic - blue

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For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

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Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

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H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

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stellarik [79]

Explanation:

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