What volume of ammonia, in #cm^3#, can be prepared by reacting 20 #cm^3# of nitrogen gas with 30 #cm^3# of hydrogen gas?

Assuming that all volume are measured at the same temperature and pressure and the reaction goes to completion.

A. 20
B. 30
C. 40
D. 50

The answer given is A. 20, why?

Answer 1

The volume of ammonia that can be prepared is #"20 cm"^3#.

This appears to be a gas volume limiting reactant problem.

Gay-Lussac's Law of Combining Volumes, which states that "the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers," must therefore be applied.

First, write the chemical equation that is balanced.

The reaction's equation is

#color(white)(ll)"N"_2color(white)(l) +color(white)(l) "3H"_2 → "2NH"_3# #color(white)(m)1color(white)(mmml) 3color(white)(mmml) 2# #"1 cm"^3color(white)(m) "3 cm"^3color(white)(ml) "2 cm"^3#

The ratio is 1:3:2, according to the balanced equation's coefficients.

Thus, we may state that

#"1 cm"^3 "of N"_2 + "3 cm"^3 "of H"_2 → "2 cm"^3 "of NH"_3#

Step 2: Determine the reactant that is limiting.

From #"N"_2#: #20 color(red)(cancel(color(black)("cm"^3 "N"_2))) × ("2 cm"^3 color(white)(l)"NH"_3)/(1 color(red)(cancel(color(black)("cm"^3 "N"_2)))) = "30 cm"^3color(white)(l) "NH"_3#
From #"H"_2#: #30 color(red)(cancel(color(black)("cm"^3 "H"_2))) × ("2 cm"^3 "NH"_3)/(3 color(red)(cancel(color(black)("cm"^3 "H"_2)))) = "20 cm"^3color(white)(l) "NH"_3#
#"H"_2# is the limiting reactant, because it forms the smaller amount of #"NH"_3#.
You can't get more than #"20 cm"^3color(white)(l) "NH"_3#.
By the time you have formed that amount, all the #"H"_2# will have been used up, and you won't be able to form any more #"NH"_3#.
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Answer 2

The balanced chemical equation for the reaction between nitrogen gas (N2) and hydrogen gas (H2) to form ammonia (NH3) is:

N2(g) + 3H2(g) → 2NH3(g)

According to the stoichiometry of the reaction, 1 volume of nitrogen gas reacts with 3 volumes of hydrogen gas to produce 2 volumes of ammonia gas.

Given that 20 cm^3 of nitrogen gas and 30 cm^3 of hydrogen gas are reacted, we can determine the volume of ammonia gas produced using the volume ratios from the balanced equation.

20 cm^3 of N2 will require (3/1) * 20 = 60 cm^3 of H2 to react completely.

Since only 30 cm^3 of H2 is available, we can only react (1/3) * 30 = 10 cm^3 of N2.

Therefore, the limiting reactant is hydrogen gas, and according to the stoichiometry, it will produce half its volume in ammonia gas. Thus, the volume of ammonia gas produced is 10 cm^3 / 2 = 5 cm^3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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