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3 years ago

6

Ammonia (nh3) is widely used as a fertilizer and in many household cleaners. how much ammonia is produced when 6.64 mol of hydro

gen gas react with an excess of nitrogen gas

1 answer:

olga55 [171]3 years ago

5 0

N₂ + 3H₂ ⇒ 2NH₃

doesnt matterN₂ + 6.64H₂ ⇒ 2NH₃

(6.64H₂/3H₂) x (2NH₃) =4.4266667

rounded to sig figs= 4.43

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First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

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So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

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As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

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