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
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
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
To determine the balanced equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O), we need to follow these steps:
Step 1: Write the chemical formula for each compound involved in the reaction:
Ammonia = NH3
Oxygen = O2
Nitrogen monoxide = NO
Water = H2O
Step 2: Assign coefficients to balance the equation:
NH3 + O2 -> NO + H2O
Step 3: Determine the limiting reagent:
To identify the limiting reagent, we need to compare the moles of each reactant to their stoichiometric coefficients in the balanced equation.
The stoichiometric coefficients in the balanced equation are:
NH3: coefficient 1
O2: coefficient 1
NO: coefficient 1
H2O: coefficient 1
Given:
Moles of NH3 = 2.36 mol
Moles of O2 = 6.30 mol
Based on the coefficients, both NH3 and O2 have equal stoichiometric coefficients, which means they have a 1:1 mole ratio in the reaction. Therefore, neither NH3 nor O2 is in excess, and both will react completely.
Step 4: Use the balanced equation to determine the amount of product formed:
Since NH3 and O2 react in a 1:1 ratio to form NO and H2O, the amount of NO and H2O formed will also be 2.36 mol.
Therefore, the amount of NO and H2O formed when 2.36 mol of NH3 and 6.30 mol of O2 react is:
amount of NO = 2.36 mol
amount of H2O = 2.36 mol
To determine the balanced equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O), we need to follow the steps for balancing equations:
1. Write the unbalanced equation: NH3 + O2 -> NO + H2O
2. Count the number of atoms on each side of the equation:
On the left side:
- Nitrogen: 1
- Hydrogen: 3
- Oxygen: 2
On the right side:
- Nitrogen: 1
- Hydrogen: 2
- Oxygen: 3
3. Start with the most complex molecule and adjust the coefficients to balance the atoms. In this case, we can start with NH3 and balance the nitrogen and hydrogen atoms first:
NH3 + O2 -> N(NO) + H2O
4. Next, balance the remaining oxygen atoms. Since the only oxygen-containing molecule on the right side is H2O, we can balance the equation as follows:
4NH3 + 5O2 -> 4N(NO) + 6H2O
Now that we have a balanced equation, we can determine the amount of NO and H2O formed when 2.36 mol NH3 and 6.30 mol O2 react. Here's how we can approach this problem using an ICF table:
1. Write the balanced equation:
4NH3 + 5O2 -> 4N(NO) + 6H2O
2. Set up an ICF (Initial, Change, Final) table:
| NH3 | O2 | N(NO) | H2O
--------------------------------------------
Initial amount | 2.36 | 6.30 | 0 | 0
Change in amount | -4x | -5x | +4x | +6x
Final amount | 2.36 - 4x | 6.30 - 5x | 4x | 6x
3. Determine the limiting reagent by comparing the number of moles of NH3 and O2:
Divide the initial amount of each reactant by its coefficient in the balanced equation:
NH3: (2.36 mol) / 4 = 0.59 mol
O2: (6.30 mol) / 5 = 1.26 mol
The reactant with the smaller number of moles is the limiting reagent, which is NH3 in this case. Therefore, NH3 determines the amount of product formed.
4. Use the moles of the limiting reagent (NH3) to determine the moles of the products:
From the ICF table, we can see that the final amount of N(NO) is 4x and the final amount of H2O is 6x.
Since the balanced equation shows that 4 moles of NH3 react to form 4 moles of N(NO) and 6 moles of NH3 react to form 6 moles of H2O, we can conclude that the moles of the products are equal to the moles of the limiting reagent.
Therefore, the amount of NO and H2O formed when 2.36 mol NH3 and 6.30 mol O2 react is:
amount of NO = 4x = 4(0.59 mol) = 2.36 mol
amount of H2O = 6x = 6(0.59 mol) = 3.54 mol
Hence, the amounts of NO and H2O formed when 2.36 mol NH3 and 6.30 mol O2 react are 2.36 mol and 3.54 mol, respectively.