How many grams if nh3 can be produced from 2.57mol of n2 and excess h2?
2.57 mol of N2 --> 2 * 14 * 2.57 = 72 g of N2
that is 5.14 mol of N so 3*5.14 = 15.4 mol of H which is 1 g/mol
15.4 mol * 1 = 15.4 g of H
so
72 + 15.4 = 87.4 grams of NH3
Well, let me do some math and then add a little joke to sweeten the deal.
First, we need to look at the balanced chemical equation for the reaction between N2 and H2 to produce NH3:
N2 + 3H2 -> 2NH3
From this equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
So, if we have 2.57 moles of N2, we can use stoichiometry to find the number of moles of NH3 produced:
2.57 moles N2 * (2 moles NH3 / 1 mole N2) = 5.14 moles NH3
Finally, to convert moles to grams, we can use the molar mass of NH3, which is approximately 17.03 g/mol:
5.14 moles NH3 * 17.03 g/mol = 87.6 grams of NH3.
And there you have it! Approximately 87.6 grams of NH3 can be produced from 2.57 moles of N2. Now, here's a joke to lighten the mood:
Why was the math book sad?
Because it had too many problems!
To determine the number of grams of NH3 that can be produced from a given amount of N2, we need to use the balanced chemical equation for the reaction between N2 and H2 to produce NH3.
The balanced chemical equation is as follows:
N2 + 3H2 -> 2NH3
From the equation, we can see that for every 1 mole of N2, we produce 2 moles of NH3. Therefore, the ratio of N2 to NH3 is 1:2.
Given that you have 2.57 moles of N2, we can use this ratio to calculate the moles of NH3 that can be produced:
Moles of NH3 = Moles of N2 * (2/1)
Moles of NH3 = 2.57 mol * (2/1) = 5.14 mol
Now, to find the grams of NH3, we need to multiply the moles of NH3 by its molar mass. The molar mass of NH3 is calculated as follows:
Molar mass of NH3 = (1 x atomic mass of N) + (3 x atomic mass of H)
= (1 x 14.01 g/mol) + (3 x 1.01 g/mol)
= 14.01 g/mol + 3.03 g/mol
= 17.04 g/mol
Finally, we can calculate the grams of NH3:
Grams of NH3 = Moles of NH3 * Molar mass of NH3
= 5.14 mol * 17.04 g/mol
= 87.74 g
Therefore, 2.57 mol of N2 can produce 87.74 grams of NH3 when excess H2 is present.
To determine the number of grams of NH3 produced, you need to follow a series of steps. Here's a step-by-step guide on how to calculate it:
Step 1: Write a balanced equation for the reaction between N2 and H2 to produce NH3.
N2 + 3H2 → 2NH3
Step 2: Determine the molar ratio between N2 and NH3 from the balanced equation.
From the equation, we see that 1 mole of N2 reacts with 2 moles of NH3.
Step 3: Convert the given amount of N2 in moles to moles of NH3.
Using the molar ratio obtained in Step 2, we can calculate the moles of NH3 produced from 2.57 moles of N2.
Moles NH3 = Moles N2 × (2 moles NH3 / 1 mole N2)
Moles NH3 = 2.57 mol N2 × (2 mol NH3 / 1 mol N2) = 5.14 mol NH3
Step 4: Convert moles of NH3 to grams using its molar mass.
The molar mass of NH3 can be calculated by summing the atomic masses of nitrogen (N) and hydrogen (H).
Molar mass of NH3 = (1 × atomic mass of N) + (3 × atomic mass of H)
Molar mass of NH3 = (1 × 14.01 g/mol) + (3 × 1.01 g/mol) = 17.03 g/mol
To determine the mass of NH3 produced, multiply the moles of NH3 obtained in Step 3 by the molar mass of NH3.
Mass of NH3 = Moles NH3 × Molar mass of NH3
Mass of NH3 = 5.14 mol NH3 × 17.03 g/mol = 87.60 g
Therefore, approximately 87.60 grams of NH3 can be produced from 2.57 moles of N2 and excess H2.