Ammonia is produced by the reaction of hydrogen and nitrogen.
N2(g)+3H2(g)→2NH3(g)ammonia
How many moles of N2 reacted if 0.65mole NH3 is produced?
How many moles of NH3 are produced when 1.2moles H2 reacts?
To determine how many moles of N2 reacted if 0.65 moles of NH3 is produced, you need to use the balanced equation and stoichiometry.
According to the balanced equation:
N2(g) + 3H2(g) → 2NH3(g)
The stoichiometric coefficient of N2 tells us that for every 1 mole of N2, 2 moles of NH3 will be produced.
So, you can set up a ratio using this information:
1 mole N2 : 2 moles NH3
Let x be the number of moles of N2 reacted.
Using the ratio, you can set up the following equation:
1 mole N2 / 2 moles NH3 = x moles N2 / 0.65 moles NH3
Simplifying this equation, you get:
x = (1 mole N2 / 2 moles NH3) * (0.65 moles NH3) = 0.325 moles N2
Therefore, 0.325 moles of N2 reacted if 0.65 moles of NH3 is produced.
To determine how many moles of NH3 are produced when 1.2 moles of H2 reacts, you can use the balanced equation and stoichiometry.
According to the balanced equation:
N2(g) + 3H2(g) → 2NH3(g)
The stoichiometric coefficient of H2 tells us that for every 3 moles of H2, 2 moles of NH3 will be produced.
So, you can set up a ratio using this information:
3 moles H2 : 2 moles NH3
Let y be the number of moles of NH3 produced.
Using the ratio, you can set up the following equation:
3 moles H2 / 2 moles NH3 = 1.2 moles H2 / y moles NH3
Simplifying this equation, you get:
y = (2 moles NH3 / 3 moles H2) * (1.2 moles H2) = 0.8 moles NH3
Therefore, 0.8 moles of NH3 are produced when 1.2 moles of H2 reacts.
To find the number of moles of N2 that reacted when 0.65 moles of NH3 is produced, we can use the stoichiometry of the reaction.
From the balanced chemical equation, we can see that 1 mole of N2 reacts to produce 2 moles of NH3.
So, if x moles of N2 reacts, it will produce (2x) moles of NH3.
Now, we can set up a proportion to solve for x:
(2x moles NH3 / 1 mole N2) = (0.65 moles NH3 / x mole N2)
Cross-multiplying the proportion, we get:
2x = 0.65
Dividing both sides by 2, we find:
x = 0.65 / 2
x = 0.325 moles of N2
Therefore, 0.325 moles of N2 reacted to produce 0.65 moles of NH3.
To find the number of moles of NH3 produced when 1.2 moles of H2 reacts, we can again use the stoichiometry of the reaction.
From the balanced chemical equation, we can see that 3 moles of H2 react to produce 2 moles of NH3.
So, if y moles of H2 react, it will produce (2/3 * y) moles of NH3.
Now, we can set up a proportion to solve for y:
((2/3) * y moles NH3 / 3 moles H2) = (x moles NH3 / 1.2 moles H2)
Cross-multiplying the proportion, we get:
(2/3) * y = (1.2 * x)
Dividing both sides by (2/3), we find:
y = (1.2 * x) / (2/3)
Simplifying further, we get:
y = (1.2 * x) * (3/2)
y = (1.8 * x)
Therefore, 1.8 times the number of moles of H2 is the number of moles of NH3 produced.
So, when 1.2 moles of H2 reacts, it will produce:
(1.8 * 1.2) moles of NH3 = 2.16 moles of NH3
Remember that you can convert any material in mols in the equation to any other material in mols by using the coefficients in the balanced equation.
0.65 mole NH3 x (1 mol N2/2 mol NH3) = 0.65 x 1/2 = 0.325 mols N2 reacted and I know that's too many s.f.
Important: Note here that this is the same kind of conversion as with the mols and molecules.
a number x conversion factor. The only difference in the two set of problems is that the conversion factor in this problem is made up of the coefficients in the balanced equation while in the mole/atoms problems it was made up of the unit factor.