Based on the balanced equation
2Al + 6HCl → 2AlCl3 + 3H2
calculate the number of excess reagent units remaining when 140 Al atoms and 402 HCl molecules react?
nvm i figure out how to do this kind of problem but can u help me with my other question
I worked the second post before I looked at this one. Both are limiting reagent problems.
To determine the number of excess reagent units remaining, we need to compare the stoichiometric ratio of the reactants and products.
From the balanced equation, we know that 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl3 and 3 moles of H2.
To convert the given quantities to moles, we can use the Avogadro's number (6.022 × 10^23 molecules/mole):
140 Al atoms * (1 mole/6.022 × 10^23 atoms) = 2.322 moles of Al
402 HCl molecules * (1 mole/6.022 × 10^23 molecules) = 0.667 moles of HCl
Now let's compare the stoichiometry:
Al:HCl ratio: 2 moles : 6 moles = 1 : 3
Actual ratio: 2.322 moles : 0.667 moles ≈ 3.48 : 1
Since the actual ratio is greater than the stoichiometric ratio, we can conclude that Al is the limiting reagent.
To calculate the excess reagent remaining, we need to determine the amount of HCl that reacted.
Using the stoichiometric ratio, we can calculate the amount of HCl required to react with 2.322 moles of Al:
(2.322 moles Al) * (6 moles HCl/2 moles Al) = 6.966 moles HCl
Therefore, the excess HCl remaining is:
Actual amount of HCl - required amount of HCl = 0.667 moles - 6.966 moles ≈ -6.3 moles
The negative value indicates that there is no excess HCl remaining since it is completely consumed by the reaction with Al.
To calculate the number of excess reagent units remaining, we first need to determine the limiting reagent.
1. Start by converting the given quantities into moles.
- 140 Al atoms: The molar mass of Al is 26.98 g/mol, and since 1 mol contains 6.022 × 10^23 atoms (Avogadro's number), we can calculate the number of moles of Al by dividing the number of atoms by Avogadro's number.
Moles of Al = 140 atoms / (6.022 × 10^23 atoms/mol) = 2.324 × 10^-22 mol Al
- 402 HCl molecules: The molar mass of HCl is 36.46 g/mol, and since 1 mol contains 6.022 × 10^23 molecules (Avogadro's number), we can calculate the number of moles of HCl by dividing the number of molecules by Avogadro's number.
Moles of HCl = 402 molecules / (6.022 × 10^23 molecules/mol) = 6.672 × 10^-22 mol HCl
2. Next, write the balanced equation to determine the stoichiometric ratio between Al and HCl.
From the equation: 2Al + 6HCl → 2AlCl3 + 3H2
The ratio of Al to HCl is 2:6, or simply 1:3.
3. Compare the ratio of Al to HCl, keeping in mind the smallest value determines the limiting reagent.
Since the ratio is 1:3 and there are fewer moles of Al than HCl, Al is the limiting reagent as there are not enough Al atoms to react with all the HCl molecules.
4. Calculate the moles of HCl required to react with the limiting reagent.
Moles of HCl required = Moles of Al x (3 moles of HCl / 1 mole of Al)
= 2.324 × 10^-22 mol Al x (3 mol HCl / 1 mol Al)
= 6.972 × 10^-22 mol HCl
5. Calculate the excess reagent units remaining.
Excess reagent units remaining = Initial moles of HCl - Moles of HCl required
= 6.672 × 10^-22 mol HCl - 6.972 × 10^-22 mol HCl
= -0.300 × 10^-22 mol HCl
Since the result is negative, it indicates that there is an excess of HCl remaining after the reaction. However, the negative sign does not have any meaning in this context. So, the number of excess reagent units remaining is approximately 0.