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rusak2 [61]
3 years ago
11

For the decomposition of dinitrogen pentoxide in carbon tetrachloride solution at 30 °C , the average rate of disappearance of N

2O5 over the time period from t = 0 min to t = 112 min is found to be 5.80×10^-4 M min^-1. The average rate of formation of O2 over the same time period is:__________ M min-1
Chemistry
1 answer:
Sonbull [250]3 years ago
4 0

Answer:

1.45 × 10⁻³ M min⁻¹

Explanation:

Step 1: Write the balanced reaction for the decomposition of dinitrogen pentoxide

2 N₂O₅(g) ⇒ 2 N₂(g) + 5 O₂(g)

Step 2: Calculate the average rate of formation of oxygen

The average rate of disappearance of N₂O₅ over the time period from t = 0 min to t = 112 min is found to be 5.80 × 10⁻⁴ M min⁻¹ (5.80 × 10⁻⁴ molN₂O₅ L⁻¹  min⁻¹). Given that the molar ratio of N₂O₅ to O₂ is 2:5, the average rate of formation of O₂ is:

5.80 × 10⁻⁴ molN₂O₅ L⁻¹  min⁻¹ × 5 molO₂/ 2 molN₂O₅ = 1.45 × 10⁻³ molO₂ L⁻¹  min⁻¹ = 1.45 × 10⁻³ M min⁻¹

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then, we need to get the concentration of [Na2HPO4]:
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A solution is to be prepared with a sodium ion concentrationof
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Answer : The mass of sodium sulfate needed is 5.7085 grams.

Explanation : Given,

Concentration of sodium ion = 0.148 mol/L

Volume of solution = 2.29 L

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First we have to determine the moles of sodium ion.

\text{Concentration of sodium ion}=\frac{\text{Moles of sodium ion}}{\text{Volume of solution}}

0.184mol/L=\frac{\text{Moles of sodium ion}}{2.29L}

\text{Moles of sodium ion}=0.08035mole

Now we have to calculate the moles of sodium sulfate.

The balanced chemical reaction will be,

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As, 2 moles of sodium ion produced from 1 moles of Na_2SO_4

So, 0.08035 moles of sodium ion produced from \frac{0.08035}{2}=0.040175 moles of Na_2SO_4

Now we have to calculate the mass of sodium sulfate.

\text{Mass of }Na_2SO_4=\text{Moles of }Na_2SO_4\times \text{Molar mass of }Na_2SO_4

\text{Mass of }Na_2SO_4=0.040175mole\times 142g/mole=5.7085g

Therefore, the mass of sodium sulfate needed is 5.7085 grams.

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