Study the chemical reaction.
N2(g) + 3H2(g) → 2NH3(g)
At the end of the chemical reaction, 5 moles of NH3 are produced.
How many moles of N2 and H2 entered the reaction?
A)2 moles of N2
6 moles of H2
B)1.5 moles of N2
4.5 moles of H2
C)2.5 moles of N2
7.5 moles of H2
D)1 mole of N2
3 moles of H2
C?
yup
To determine how many moles of N2 and H2 entered the reaction, we can use the stoichiometry of the balanced chemical equation.
The balanced equation is:
N2(g) + 3H2(g) → 2NH3(g)
According to the stoichiometry, the ratio of moles of N2 to moles of NH3 is 1:2, and the ratio of moles of H2 to moles of NH3 is 3:2.
Since 5 moles of NH3 are produced, we can calculate the moles of N2 and H2 by multiplying the moles of NH3 by the corresponding ratios:
Moles of N2 = 1/2 * 5 = 2.5 moles
Moles of H2 = 3/2 * 5 = 7.5 moles
Therefore, the correct answer is option C) 2.5 moles of N2 and 7.5 moles of H2 entered the reaction.
To determine the number of moles of N2 and H2 that entered the reaction, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is: N2(g) + 3H2(g) → 2NH3(g)
From this equation, we can see that for every 1 mole of N2, 3 moles of H2 are needed to produce 2 moles of NH3. Therefore, the ratio of moles of N2 to moles of H2 is 1:3.
Given that 5 moles of NH3 were produced at the end of the reaction, we can determine the number of moles of N2 and H2 that entered the reaction based on this ratio.
If we assume x moles of N2 entered the reaction, then according to the ratio, 3x moles of H2 would have also entered the reaction.
Since we have 5 moles of NH3 produced, and each mole of NH3 is formed from 1 mole of N2, we can set up the equation:
x/1 = 5/2
Cross-multiplying, we get:
2x = 5
Dividing both sides by 2, we find:
x = 5/2 = 2.5 moles of N2
Since 3x moles of H2 entered the reaction, we can calculate:
3 * 2.5 = 7.5 moles of H2
Therefore, the answer is option C) 2.5 moles of N2 and 7.5 moles of H2 entered the reaction.