N2+ 3H2--> 2NH3
How many grams of NH3 can be produced from the reaction of 28 g N2, and 25 g of H2??
I got 34g NH3 and 140 NH2
2. How much of the excess reagent in problem 1 is left over?
No, no, and no.
28 g N2 = ??moles = 28/14 = 2 moles N2.
2 moles N2 x (2 moles NH3/1 mole N2) = 2*2/1 = 4 moles NH3.
grams = moles NH3 x molar mass NH3 = about close to 70 g.
There is NO NH2 produced. It isn't even in the equation.
How much H2 remains un-reacted?
You have 25 g H2 which is 25/2 = 12.5 moles. How much H2 is used? That is
2 moles N2 x (3 moles H2/1 mole N2) = 2*3/1 = 6.0.
So you must have 12.5-6.0 moles left over which you can convert to grams if that is desired.
To solve this problem, we need to use the concept of the limiting reactant.
1. First, we need to determine which reactant is the limiting reactant. The limiting reactant is the one that will be completely used up before the other reactant is consumed. To do this, we need to calculate how much NH3 can be produced from both N2 and H2.
a. Calculate the number of moles of N2:
Moles of N2 = Mass of N2 / Molar mass of N2
Moles of N2 = 28 g / 28 g/mol = 1 mole of N2
b. Calculate the number of moles of H2:
Moles of H2 = Mass of H2 / Molar mass of H2
Moles of H2 = 25 g / 2 g/mol = 12.5 moles of H2
c. Use the balanced chemical equation to determine the moles of NH3 that can be produced. From the balanced equation, we see that 1 mole of N2 reacts to produce 2 moles of NH3, and 3 moles of H2 react to produce 2 moles of NH3. Therefore, we need to compare the ratio of moles of N2 and H2 to determine the limiting reactant:
Ratio N2:H2 for NH3 production = (1 mole N2) / (3 moles H2) = 1:3
The ratio shows that we need 3 moles of H2 for every 1 mole of N2 to produce NH3. Since the actual ratio of moles of H2 to N2 is 12.5:1, we can see that there is an excess of H2. Therefore, N2 is the limiting reactant.
d. Calculate the number of moles of NH3 produced from the limiting reactant:
Moles of NH3 = Moles of N2 * (2 moles NH3 / 1 mole N2) = 1 * 2 = 2 moles of NH3
e. Calculate the mass of NH3 produced:
Mass of NH3 = Moles of NH3 * Molar mass of NH3
Mass of NH3 = 2 moles * 17 g/mol = 34 g NH3
So, the correct answer is 34 g of NH3 can be produced from the given reactants of 28 g N2 and 25 g H2.
2. To find out how much of the excess reagent is left over, we need to compare the moles of the excess reagent to the moles of the limiting reactant. In this case, we know that N2 is the limiting reactant, and we calculated earlier that we need 1 mole of N2 to react with 3 moles of H2.
Using this information, we can calculate the moles of H2 required:
Moles of H2 required = (1 mole N2) * (3 moles H2 / 1 mole N2) = 3 moles of H2
Now, subtract the moles of H2 used from the total moles of H2:
Moles of H2 left over = Total moles of H2 - Moles of H2 required
Moles of H2 left over = 12.5 moles - 3 moles = 9.5 moles of H2
To find the mass of the excess H2, we can use the formula:
Mass of H2 left over = Moles of H2 left over * Molar mass of H2
Mass of H2 left over = 9.5 moles * 2 g/mol = 19 g of H2
Therefore, in problem 1, the excess reagent left over is 19 g of H2.