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borishaifa [10]
3 years ago
14

A solution of ammonia and water contains 2.10×1025 water molecules and 7.10×1024 ammonia molecules. How many total hydrogen atom

s are in this solution?
Chemistry
1 answer:
Natali5045456 [20]3 years ago
8 0

There are 6.33 × 10²⁵ hydrogen atoms in this solution in total.

<h3>Explanation</h3>
  • There are two hydrogen atoms in each water \text{H}_2\text{O} molecule.
  • There are three hydrogen atoms in each ammonia \text{NH}_3 molecule.

2.10 × 10²⁵ water molecules and 7.10 × 10²⁴ ammonia molecules will contain

2 \times 2.10 \times 10^{25} + 3 \times 7.10 \times 10^{24} = 6.33 \times 10^{25} hydrogen atoms in total.

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2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
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Kc = 3.90

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CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

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