calculate number of moles of NH3 if 8g of H2 reacts in excess N2
N2 + 3H2 ==> 2NH3
mols H2 = grams/molar mass = ?
Convert mols H2 to mols NH using the coefficients in the balanced equation. That's ?mols H2 x (2 mols NH3/3 mols H2) = ?
Then convert mols NH3 to grams. g = mols x molar mass = ?
To calculate the number of moles of NH3 formed, we need to use the balanced chemical equation for the reaction between H2 and N2 to produce NH3:
N2 + 3H2 -> 2NH3
According to the equation, for every 3 moles of H2, we get 2 moles of NH3.
First, we need to convert the given mass of H2 (8g) to moles using its molar mass. The molar mass of H2 is 2 g/mol (1 mole of H2 weighs 2 grams).
Moles of H2 = Mass of H2 / Molar mass of H2
Moles of H2 = 8g / 2 g/mol
Moles of H2 = 4 mol
Since the reaction is stated to have an excess of N2, we can assume that all 4 moles of H2 will react completely. Using the stoichiometric ratio from the balanced equation, we know that for every 3 moles of H2, we obtain 2 moles of NH3.
Moles of NH3 = (Moles of H2 / 3) * 2
Moles of NH3 = (4 mol / 3) * 2
Moles of NH3 ≈ 2.67 mol
Therefore, approximately 2.67 moles of NH3 will be formed.
To calculate the number of moles of NH3 produced when 8g of H2 reacts in excess N2, we need to use the concept of stoichiometry. The balanced equation for the reaction between H2 and N2 to produce NH3 is:
N2 + 3H2 -> 2NH3
From the equation, we can see that 3 moles of H2 are needed to produce 2 moles of NH3. Therefore, we can set up a proportion to find the number of moles of NH3 produced:
(2 moles NH3 / 3 moles H2) = (x moles NH3 / 8g H2)
Cross-multiplying and solving for x gives us:
x = (2 moles NH3 / 3 moles H2) * 8g H2
x = (2/3) * 8g
x = 16/3 g
To convert this to moles, we divide by the molar mass of NH3, which is approximately 17g/mol:
Number of moles of NH3 = (16/3 g) / (17 g/mol)
Number of moles of NH3 = 0.314 moles
Therefore, when 8g of H2 reacts in excess N2, approximately 0.314 moles of NH3 are produced.