4NH3 + 5O2 --> 4NO + 6H2O
170g of ammonia (NH3) are allowed to react with 384g of O2.
how many moles of NO is produced? and also how many grams of NO will be produced?
figure the moles of NH3 in 170g.
now figure the moles of O2 in 384g
if you have 5/4 more moles of O2 than NH3, then NH3 is the limiting reageant. Moles NO=molesNH3*
if you have less than 5/4 moles of O2, then O2 is limiting. Moles NO = 4/5 moles of O2 reacted.
To find the number of moles of NO produced, we need to use the balanced chemical equation and the given masses of ammonia (NH3) and oxygen (O2).
The molar mass of NH3 is:
NH3 = 1(N) + 3(H) = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol
The molar mass of O2 is:
O2 = 2(O) = 2(16.00 g/mol) = 32.00 g/mol
Now, we can calculate the number of moles of each reactant:
Moles of NH3 = Mass of NH3 / Molar mass of NH3
Moles of NH3 = 170 g / 17.03 g/mol = 9.987 moles (approximately 10 moles)
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 384 g / 32.00 g/mol = 12 moles
From the balanced equation, we can see that the mole ratio between NH3 and NO is 4:4 or 1:1. Therefore, the number of moles of NO produced is also 10 moles.
To calculate the mass of NO produced, we need to use the molar mass of NO. The molar mass of NO is:
NO = 1(N) + 1(O) = 14.01 g/mol +16.00 g/mol = 30.01 g/mol
Mass of NO = Number of moles of NO x Molar mass of NO
Mass of NO = 10 moles x 30.01 g/mol = 300.1 g (approximately 300 g)
Therefore, 10 moles of NO and approximately 300 grams of NO will be produced in the given reaction.
To determine the number of moles of NO (nitric oxide) produced, you need to use the balanced chemical equation and the given quantities of reactants.
1. Calculate the number of moles of NH3:
To find the number of moles, divide the given mass by the molar mass of NH3.
Molar mass of NH3 = 14.01 g/mol (N) + 3(1.01 g/mol) (H) = 17.03 g/mol
Number of moles of NH3 = 170 g / 17.03 g/mol = 9.99 mol (approx.)
2. Calculate the number of moles of O2:
Similarly, divide the given mass by the molar mass of O2.
Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol
Number of moles of O2 = 384 g / 32.00 g/mol = 12.00 mol
3. Determine the limiting reactant:
The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed. To find it, compare the moles of NH3 and O2. According to the balanced equation, the mole ratio between NH3 and O2 is 4:5. Therefore, the ratio of moles is also 4:5. Since the ratio of NH3 to O2 is smaller than this, NH3 is the limiting reactant.
4. Calculate the moles of NO produced:
Using the mole ratio from the balanced equation, the ratio of NH3 to NO is 4:4. Therefore, the moles of NO produced are the same as the moles of NH3, which is 9.99 mol.
5. Calculate the grams of NO produced:
To find the grams of NO produced, multiply the moles of NO by the molar mass of NO.
Molar mass of NO = 14.01 g/mol (N) + 16.00 g/mol (O) = 30.01 g/mol
Grams of NO produced = 9.99 mol * 30.01 g/mol = 299.70 g (approx.)
Therefore, approximately 9.99 moles of NO and 299.70 grams of NO will be produced.