Al2O3(s)+6NO(g)---->5N2(g)+6H2O(l)
how many grams of Nitric Oxide are required to produce 145g of nitrogen in the following reaction?
as written, 6 moles of NO produce 5 moles of N2
(The equation doesn't balance, but I assume there are other products involved - ?)
mass of N2 = 28
145g NO = 145/28 = 5.18 moles
5.18mN2 * 6mNO/5mN2 = 6.21mNO
mass of NO = 30
6.21m * 30g/m = 186g
To determine the number of grams of Nitric Oxide (NO) required to produce 145g of nitrogen (N2), we need to use stoichiometry.
From the balanced chemical equation:
Al2O3(s) + 6NO(g) -> 5N2(g) + 6H2O(l)
We can see that 6 moles of nitric oxide (NO) react to produce 5 moles of nitrogen (N2).
1. Calculate the molar mass of Nitric Oxide (NO):
Nitric Oxide (NO) = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
2. Convert the given mass of nitrogen (N2) to moles:
Moles of nitrogen (N2) = mass of nitrogen (N2) / molar mass of nitrogen (N2)
Moles of nitrogen (N2) = 145 g / (28.02 g/mol) = 5.177 mol
3. Use the mole ratio from the balanced equation to determine the moles of Nitric Oxide (NO) required:
Moles of Nitric Oxide (NO) = (Moles of nitrogen (N2) / 5) * 6
Moles of Nitric Oxide (NO) = (5.177 mol / 5) * 6 = 6.2124 mol
4. Convert the moles of Nitric Oxide (NO) to grams:
Mass of Nitric Oxide (NO) = Moles of Nitric Oxide (NO) * molar mass of Nitric Oxide (NO)
Mass of Nitric Oxide (NO) = 6.2124 mol * 30.01 g/mol = 186.45 g
Therefore, approximately 186.45 grams of Nitric Oxide (NO) are required to produce 145 grams of nitrogen (N2) in the given reaction.