Can someone check my first answer and help me with the second question? Thanks so much!! :D
1. N2 + 3H2 ==> 2NH3 : How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2?
==> I got 37.5 grams.
2. How much of the excess reagent in Problem 1 is left over?
==> I really have no idea how to do this. Can someone please explain this to me? Thanks!
1. N2 + 3H2 ==> 2NH3 (balanced equation)
Now put the RAM (relative atomic masses) to each reactant.
2*14 + 3*2 ==> 2*(17)
28 + 6 ==> 34
Therefore 28 grams of N2 with 6 grams of H2 will yield 34 grams of NH3 (assuming one of the reactants will be exhausted).
2.
Since only 6 grams of H2 out of 25 grams has been used for the reaction, what would be the quantity left?
Of course! I'd be happy to help explain how to solve the second question.
To find out how much excess reagent is left over in a chemical reaction, we first need to identify the limiting reagent. The limiting reagent is the reactant that is completely consumed first and determines the maximum amount of product that can be formed.
In this case, we know from Problem 1 that 28 g of N2 and 25 g of H2 were used in the reaction. To determine the limiting reagent, we need to calculate the number of moles of each reactant.
To calculate the number of moles, we can use the molar mass of each substance. The molar mass of nitrogen gas (N2) is approximately 28.02 g/mol, and the molar mass of hydrogen gas (H2) is approximately 2.02 g/mol.
Number of moles of N2 = mass of N2 / molar mass of N2
= 28 g / 28.02 g/mol
≈ 0.999 mol
Number of moles of H2 = mass of H2 / molar mass of H2
= 25 g / 2.02 g/mol
≈ 12.38 mol
Now let's look at the balanced chemical equation:
N2 + 3H2 ==> 2NH3
According to the balanced equation, the mole ratio between N2 and NH3 is 1:2. This means that for every 1 mole of N2, 2 moles of NH3 are produced.
Since we have 0.999 mol of N2, we can calculate the maximum amount of NH3 that can be produced:
Number of moles of NH3 = 2 * (moles of N2)
= 2 * 0.999 mol
≈ 1.998 mol
To convert this into grams, we can use the molar mass of NH3, which is approximately 17.03 g/mol.
Mass of NH3 = number of moles of NH3 * molar mass of NH3
= 1.998 mol * 17.03 g/mol
≈ 34.01 g
Therefore, the maximum amount of NH3 that can be produced from the given reactants is approximately 34.01 grams.
To find the amount of excess reagent left over, we compare the amount actually used with the amount that was required to react completely.
If 25 g of H2 were used, and the stoichiometry of the reaction states that 3 moles of H2 are required to react with 1 mole of N2, we can calculate the amount of H2 required to react completely:
Amount of H2 required = (moles of N2) / 3
= 0.999 mol / 3
≈ 0.333 mol
To convert this into grams, we can use the molar mass of H2:
Mass of H2 required = amount of H2 required * molar mass of H2
= 0.333 mol * 2.02 g/mol
≈ 0.67 g
Now, to find the excess reagent left over, we subtract the amount used from the amount required:
Excess reagent left over = amount of excess reagent initially - amount of excess reagent used
= 25 g - 0.67 g
≈ 24.33 g
Therefore, based on the calculations, approximately 24.33 grams of the excess reagent (H2) would be left over in this reaction.