x NH3 + y O2 ---> x NO + z H2O
to balance H
3 x = 2 z so z = (3/2) x
to balance O
2 y = x + z so z = 2 y - x
then
(3/2) x = 2 y - x
3 x = 4 y - 2 x
5 x = 4 y
well the least multiple I see is 20
try x = 4 and y = 5
z = (3/2) x = 6
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
that works
When solving problems involving stoichiometric coefficients, the first step is to make sure you have a balanced chemical equation. Then, you determine the limiting reagent by using the coefficients from the balanced equation. You can keep track of the amounts of all reactant and products before and after a reaction using an ICF table (as shown in the Simulation). Completing the ICF table will also allow you to determine the limiting reagent, and the amount of product formed is based on assuming that the reaction runs to completion with 100% yield. Parts A and C explore these steps in more detail. Let us consider another reaction.
Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO and H2O that would form when 2.36 mol NH3 and 6.30 mol O2 react.
Express the amounts in moles to two decimal places separated by a comma.
amount of NO, amount of H2O = ???? mol
2 answers
now
N = 14 g/mol
H = 1 g/mol
O = 16 g/mol
so
NH3 = 17 g/mol
O2 = 32 g/mol
we need 5 mol O2 for every 4 mol NH3
2.36 *5/4 = 2.95 mol O2 needed
We have excess O2 (natcherly)
so use that 2.36 mol NH3 and 2.95 mol O2'
for 2.36 mol NH3 we get 2.36 mol NO
mols H2O = (6/4) mols NH3 = 3/2 * 2.36 = 3.54 mols H2O result
N = 14 g/mol
H = 1 g/mol
O = 16 g/mol
so
NH3 = 17 g/mol
O2 = 32 g/mol
we need 5 mol O2 for every 4 mol NH3
2.36 *5/4 = 2.95 mol O2 needed
We have excess O2 (natcherly)
so use that 2.36 mol NH3 and 2.95 mol O2'
for 2.36 mol NH3 we get 2.36 mol NO
mols H2O = (6/4) mols NH3 = 3/2 * 2.36 = 3.54 mols H2O result
1 answer