How many molecules of NH3 could be generated from 64 molecules of N2 and 210 molecules of H2? How many molecules of N2 and H2 would be left over after the reaction?
To determine how many molecules of NH3 could be generated and how many molecules of N2 and H2 would be left over in the reaction, we need to consider the balanced chemical equation for the reaction. In this case, the reaction is the synthesis of ammonia (NH3) from nitrogen gas (N2) and hydrogen gas (H2), and it is represented by the following equation:
N2 + 3H2 -> 2NH3
According to the balanced equation, 1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3. Therefore, we can use this stoichiometric ratio to determine the quantities of reactants and products.
Let's start with the given amounts:
- 64 molecules of N2
- 210 molecules of H2
First, let's determine how many moles of NH3 can be generated.
Since we have 64 molecules of N2, we can convert this quantity to moles using the molar mass of N2, which is approximately 28 grams/mole.
64 molecules N2 * (1 mole N2 / 6.02 x 10^23 molecules N2) = (64/6.02 x 10^23) moles N2
Next, we need to determine the limiting reactant. The limiting reactant is the one that will be completely consumed and will limit the amount of product formed. To find the limiting reactant, we compare the ratios of moles of N2 and H2 to the stoichiometric ratio in the balanced equation.
From the balanced equation, we know that 1 mole of N2 reacts with 3 moles of H2. So, in order for 1 mole of N2 to react completely, we need 3 moles of H2.
Based on the given amounts:
- Moles of N2 = (64/6.02 x 10^23) moles N2
- Moles of H2 = 210 molecules H2 * (1 mole H2 / 6.02 x 10^23 molecules H2) = (210/6.02 x 10^23) moles H2
Now, let's calculate the moles of NH3 that can be generated. Since the stoichiometric ratio is 1:3 for N2:H2, we can use the moles of N2 to determine the moles of NH3.
Moles of NH3 = (64/6.02 x 10^23) moles N 2 * (2 moles NH3 / 1 mole N2) = (128/6.02 x 10^23) moles NH3
Finally, we can convert the moles of NH3 to molecules of NH3 using Avogadro's number of 6.02 x 10^23 molecules/mole.
Molecules of NH3 = (128/6.02 x 10^23) moles NH3 * (6.02 x 10^23 molecules NH3 / 1 mole NH3) = 128 molecules NH3
So, based on the given amounts, 128 molecules of NH3 can be generated.
To calculate the number of molecules of N2 and H2 left over after the reaction, we need to determine the excess amount of each reactant. The excess reactant is the one that is left over after the limiting reactant has been completely consumed.
From the stoichiometry, we know that for every 1 mole of N2, we need 3 moles of H2. So, the moles of H2 required to react with the given N2 can be calculated as:
Moles of H2 needed = (64/6.02 x 10^23) moles N2 * (3 moles H2 / 1 mole N2) = (192/6.02 x 10^23) moles H2
Since we have (210/6.02 x 10^23) moles H2, we can subtract the moles needed from the moles given to determine the excess moles of H2.
Excess moles of H2 = (210/6.02 x 10^23) moles H2 - (192/6.02 x 10^23) moles H2
Similarly, you can calculate the excess moles of N2 by comparing the moles of N2 provided to the calculated moles needed for the reaction.
Excess moles of N2 = (64/6.02 x 10^23) moles N2 - (64/6.02 x 10^23) moles N2
Finally, to convert the excess moles of H2 and N2 back to molecules, use Avogadro's number:
Excess molecules of H2 = Excess moles of H2 * (6.02 x 10^23 molecules H2 / 1 mole H2)
Excess molecules of N2 = Excess moles of N2 * (6.02 x 10^23 molecules N2 / 1 mole N2)
By performing the calculations above, you can determine the number of molecules of N2 and H2 that would be left over after the reaction.