25 g of NH3 is mixed with 4 moles of O2 in the given reaction: 4NH3(g)+5O2(g) revers reaction 4NO(g)+6H2O(1) whati mass of NO is formed?
First, we need to determine which reactant is limiting. To do this, we can convert the mass of NH3 to moles:
25 g NH3 x (1 mol NH3/17.03 g NH3) = 1.47 mol NH3
Next, we use the stoichiometry of the reaction to determine how many moles of NO will be produced for this amount of NH3:
1.47 mol NH3 x (4 mol NO/4 mol NH3) = 1.47 mol NO
So, 1.47 moles of NO will be produced. To convert this to mass, we can use the molar mass of NO:
1.47 mol NO x (30.01 g NO/1 mol NO) = 44.1 g NO
Therefore, 44.1 g of NO will be formed.
To find the mass of NO formed, we first need to determine the limiting reactant. The limiting reactant is the one that is completely consumed, thereby determining the maximum amount of product that can be formed.
Let's start by calculating the number of moles of NH3 and O2:
Moles of NH3 = mass / molar mass
Molar mass of NH3 (NH3) = 1(1) + 3(1) = 17 g/mol (1 nitrogen + 3 hydrogens)
Moles of NH3 = 25 g / 17 g/mol = 1.47 moles
Moles of O2 = 4 moles (given)
Now, let's compare the moles of NH3 and O2 using the stoichiometry of the reaction:
4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(l)
From the balanced equation, we can see that the ratio of NH3 to NO is 4:4 (1:1) and the ratio of O2 to NO is 5:4. Therefore, the ratio of NH3 to O2 to NO is 4:5:4.
Since O2 has a smaller ratio compared to NH3, it is the limiting reactant. This means that NH3 is in excess, but O2 is the reactant that determines the amount of product formed.
To find the moles of NO formed, we use the ratio of O2 to NO in the balanced equation:
Moles of NO = (Moles of O2) * (4 moles of NO / 5 moles of O2)
= (4 mol) * (4/5)
= 3.2 moles
Finally, to find the mass of NO formed, we multiply the moles of NO by its molar mass:
Molar mass of NO (NO) = 14 + 16 = 30 g/mol (14 for nitrogen + 16 for oxygen)
Mass of NO = Moles of NO * Molar mass of NO
= 3.2 mol * 30 g/mol
= 96 g
Therefore, the mass of NO formed is 96 grams.