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2 years ago

2 NH3 + 3 CuO →3 Cu + N₂ + 3 H₂O

Chemistry

1 answer:

Tatiana [17]2 years ago

5 0

40.1g of nitrogen gas is produced.

The equation given is

2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O

This equation is already balanced.

When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.

We get 1 mole of nitrogen from 3 moles of copper oxide.

We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.

4.3/3 x 1 = 1.433 mols

1.433 mols of nitrogen gas are produced
The molar mass of nitrogen gas is 14+14 = 28g
The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g

40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.

Learn more about molarity here:

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Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

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$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

$\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}$

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

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