If 3.78 L of nitrogen gas and 28.73 L of hydrogen gas were allowed to react, how many litres of ammonia gas could form?

Volume of ammonia formed is 7.56 L

Explanation:

Volume of 1 mol of gas is 24 L.

No. of mol in 3.78 L of nitrogen =

No. of mol in 3.78 L of hydrogen =

I mol of nitrogen required 3 moles of hhydrogen.

So, 0.1575 mol of nitrogen will require 0.4725 moles of hydrogen.

As moles of hydrogen is 1.197, therefore, nitrogen will be the limiting reagent.

So,

1 mol of nitrogen forms 2 moles of ammonia

0.1575 mol of nitrogen will form 0.1575 mol × 2 = 0.315 mol of ammonia

Volume of ammonia = 0.315 × 24

                               = 7.56 L

Volume of ammonia formed is 7.56 L

that tool would be a compass

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24