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?

Answer 1

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.