Al2(SO3)3 + 6NaOH ----> 3Na2SO3 = 2 Al9OH)3
If 10.0g of Al2(SO3)3 is reacted with 10.0g of NaOH, determine the limiting reagent?
help!!!
I do these the long way.
Moles Al2*SO4)3 = grams/molar mass = ?
mols NaOH = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Al2(SO3)3 to mols Na2SO3.
Do the same for mols NaOH to mols Na2SO3.
I think the answers will not be the same which means one is wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
To determine the limiting reagent in a chemical reaction, we need to compare the amount of each reactant given and find out which one will be completely consumed first, thus limiting the amount of product that can be formed.
Let's start by calculating the number of moles of each reactant provided:
1. Aluminum sulfite (Al2(SO3)3):
- Given mass: 10.0g
- Molar mass of Al2(SO3)3:
- Atomic mass of Al = 26.98g/mol
- Atomic mass of S = 32.07g/mol
- Atomic mass of O = 16.00g/mol
- Molar mass of Al2(SO3)3 = (2 * 26.98g/mol) + (3 * (32.07g/mol + 3 * 16.00g/mol)) = 294.05g/mol
- Moles of Al2(SO3)3 = (Given mass / Molar mass) = (10.0g / 294.05g/mol) = 0.034mol
2. Sodium hydroxide (NaOH):
- Given mass: 10.0g
- Molar mass of NaOH:
- Atomic mass of Na = 22.99g/mol
- Atomic mass of O = 16.00g/mol
- Atomic mass of H = 1.01g/mol
- Molar mass of NaOH = (22.99g/mol + 16.00g/mol + 1.01g/mol) = 39.00g/mol
- Moles of NaOH = (Given mass / Molar mass) = (10.0g / 39.00g/mol) = 0.256mol
Now, let's examine the stoichiometric ratio of the reactants to determine which one is the limiting reagent. According to the balanced equation:
Al2(SO3)3 + 6NaOH -> 3Na2SO3 + 2Al(OH)3
The stoichiometric ratio between Al2(SO3)3 and NaOH is 1:6. For every 1 mole of Al2(SO3)3, we need 6 moles of NaOH.
To determine the limiting reagent, we compare the moles of each reactant with their stoichiometric ratio:
- Al2(SO3)3: NaOH = 0.034mol : 0.256mol
Based on the ratio, it shows that we have an excess amount of NaOH. There is not enough Al2(SO3)3 to react with all the NaOH completely.
Therefore, the limiting reagent is Al2(SO3)3.