How many grams of NH3 can be produced from 2.89mol of N2 and excess H2.
Express your answer numerically in grams.
N2 + 3H2 ==> 2NH3
mols N2 = 2.89
Convert to mols NH3 using the coefficients in the balanced equation.
2.89 x (2 mols NH3/1 mols N2) = 2.89 x 2/1 = ?
Then g NH3 = mols NH3 x molar mass NH3 = ?
9.5
To determine the number of grams of NH3 produced, we first need to calculate the limiting reactant between N2 and H2.
The balanced chemical equation for the reaction between N2 and H2 to produce NH3 is:
N2 + 3H2 -> 2NH3
According to the stoichiometry of the balanced equation, for every 1 mole of N2, 2 moles of NH3 are produced.
So, if we have 2.89 moles of N2, we can find the moles of NH3 produced by multiplying the moles of N2 by the mole ratio:
Moles of NH3 = 2.89 mol N2 x (2 mol NH3 / 1 mol N2)
Moles of NH3 = 5.78 mol NH3
Next, we need to convert the moles of NH3 to grams using the molar mass of NH3.
Molar mass of NH3 = 14.01 g/mol + (1.01 g/mol x 3) = 17.03 g/mol
Now, we can calculate the grams of NH3:
Grams of NH3 = 5.78 mol NH3 x 17.03 g/mol
Grams of NH3 = 98.59 g
Therefore, approximately 98.59 grams of NH3 can be produced from 2.89 moles of N2 and excess H2.
To determine the grams of NH3 produced, we need to use the balanced chemical equation for the reaction between N2 and H2 to form NH3.
The balanced chemical equation is:
N2 + 3H2 -> 2NH3
From the equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
First, we need to calculate the number of moles of NH3 produced from 2.89 moles of N2.
According to the balanced equation, the mole ratio between N2 and NH3 is 1:2. Therefore, the number of moles of NH3 produced will be:
2.89 moles N2 x (2 moles NH3 / 1 mole N2) = 5.78 moles NH3
Next, we need to convert the moles of NH3 to grams. To do this, we need to know the molar mass of NH3, which is approximately 17.03 g/mol.
To calculate the grams of NH3 produced, we can use the following formula:
Grams = Moles x Molar Mass
Grams of NH3 = 5.78 moles NH3 x 17.03 g/mol = 98.5294 g
Therefore, approximately 98.53 grams of NH3 can be produced from 2.89 moles of N2 and excess H2.