ammonia is produced by sythesizing nitrogen and hydrogen gas. if i have 2 mols of nitrogen gas and 5 mols of hydrogen gas, which reagent will run out first??
a) if i have 0.0345 mols of nitrogen gas and 0.0537 mols of hydrogen gas, which reagent will run out first? how many mols of excess reagent will have left over?
b) if i have 24g of nitrogen gas and 24g of hydrogen gas, which reagent will run out first?
There is a shorter way of doing this but I like the long way.
Convert moles each to the product, then take the smaller product amount as coming from the limiting reagent.
2 moles N2 x (2 moles NH3/2 moles N2) = 2 moles NH3.(The fractions comes from the coefficients.)
5 moles H2 x (2 moles NH3/3 moles H2) = 3.33 moles NH3.
The smaller number is N2, that is the limiting reagent.
The others are done the same way. For excess reagent, use the limiting reagent, convert to the excess reagent with the appropriate fraction to determine how much is used in the reaction, subtract that from the moles initially there to determine the amount that did not react.
hey can u show me the shorter way to do plzzz and thx for the answer... :)
To determine which reagent will run out first, we need to use the balanced chemical equation for the synthesis of ammonia:
N2 + 3H2 -> 2NH3
a) If we have 2 mols of nitrogen gas (N2) and 5 mols of hydrogen gas (H2), we can compare the molar ratio between the two reagents:
N2:H2 = 1:3
According to the chemical equation, for every 1 mol of nitrogen gas, we need 3 mols of hydrogen gas. The ratio tells us that we need a 3:1 ratio of hydrogen to nitrogen. However, we only have a 2:5 ratio, which means that hydrogen is in excess. Hence, nitrogen gas will run out first.
b) If we have 0.0345 mols of nitrogen gas and 0.0537 mols of hydrogen gas, we can still compare the molar ratio:
N2:H2 = 1:3
In this case, we have a 1:3 ratio of nitrogen to hydrogen. Since the ratio matches the stoichiometric ratio in the chemical equation, we can conclude that both reagents will be completely consumed, and there will be no excess reagent left over.
c) To determine which reagent will run out first when given masses instead of moles, we need to convert the masses of nitrogen gas and hydrogen gas to mols. To do this, we use the molar mass of each element.
The molar mass of nitrogen (N2) is 28 g/mol, and the molar mass of hydrogen (H2) is 2 g/mol.
Using the given masses:
- 24 g of nitrogen gas is equal to 24 g / 28 g/mol = 0.857 mol of nitrogen gas
- 24 g of hydrogen gas is equal to 24 g / 2 g/mol = 12 mol of hydrogen gas
Comparing the molar ratio:
N2:H2 = 1:3
According to the chemical equation, we need a 3:1 ratio of hydrogen to nitrogen. However, we have a 12:0.857 ratio, which means that nitrogen gas is in excess. Hence, hydrogen gas will run out first.