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Ostrovityanka [42]
3 years ago
9

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was fo

und that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia
Chemistry
1 answer:
Karolina [17]3 years ago
8 0

Answer:

0.327 M

Explanation:

Step 1: Write the balanced equation

2 NH₃(g) ⇄ N₂(g) + 3H₂(g)

Step 2: Make an ICE chart

        2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)

I              x             0            0

C          -2y            +y         +3y

E         x-2y             y           3y

Step 3: Find the value of y

The concentration of N₂ at equilibrium is 0.0161 M. Then,

y = 0.0161

Step 4: Find the value of x

The concentration of NH₃ at equilibrium is 0.295 M. Then,

x-2y = 0.295

x-2(0.0161) = 0.295

x = 0.327

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A 10.5 mL sample of vinegar, containing acetic acid, was titrated using 0.460 M NaOH solution. The titration required 19.13 mL o
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Explanation:

Step 1:

A good first step for a problem like this is to write down the chemical formula and balance it.

It appears here that we have 10.5 mL of vinegar, which IS acetic acid, and 19.13 mL of 0.460 M NaOH. That will give us the following balanced chemical equation:

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All of the constituents come out to a value of 1, conveniently.

Step 2:

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M stands for Molarity and V stands for volume. 1 and 2 being the before the reaction and after the reaction.

So, our M1 for this is going to be what the question says was used for this titration. That's 0.460M NaOH.

Our V1 is going to be the initial volume of the sample, which was 10.5 mL

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It's clear that we don't know M2, so let's find it.

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Using some algebra, we can see that we now have:

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