If we combined 18.00 g of KOH and 18.00 g of H2SO4 , and they reacted according to the equation below, how many moles of K2SO4 should be made if all of the KOH is used?
2 KOH + H2SO4 = 2 H2O + K2SO4
mols KOH = g/molar mass = 18.00/56.1 = 0.321
0.321 mols KOH x ( 1 mol K2SO4/2 mols KOH) = 0.321/2 = ? mols K2SO4. This assumes KOH is the limiting reagent.
To determine the number of moles of K2SO4 that will be produced when all of the KOH is used, we need to follow these steps:
Step 1: Calculate the molar mass of KOH.
The molar mass of K is 39.10 g/mol (from the periodic table).
The molar mass of O is 16.00 g/mol (from the periodic table).
The molar mass of H is 1.01 g/mol (from the periodic table).
Molar mass of KOH = (39.10 g/mol) + (16.00 g/mol) + (1.01 g/mol)
= 56.11 g/mol
Step 2: Calculate the number of moles of KOH.
moles of KOH = mass of KOH / molar mass of KOH
= 18.00 g / 56.11 g/mol
≈ 0.32 mol
Step 3: Use stoichiometry to find the number of moles of K2SO4.
From the balanced equation, we can see that 2 moles of KOH react to produce 1 mole of K2SO4.
Therefore, the number of moles of K2SO4 would be half the number of moles of KOH:
moles of K2SO4 = moles of KOH / 2
= 0.32 mol / 2
= 0.16 mol
So, 0.16 moles of K2SO4 should be produced if all of the KOH is used.
To determine how many moles of K2SO4 should be made, we need to use the given masses of KOH and H2SO4 and convert them into moles using their molar masses.
First, let's calculate the molar masses of KOH and H2SO4:
- KOH:
- K (potassium) = 39.10 g/mol
- O (oxygen) = 16.00 g/mol
- H (hydrogen) = 1.01 g/mol
- Molar mass = 39.10 g/mol + 16.00 g/mol + 1.01 g/mol = 56.11 g/mol
- H2SO4:
- H (hydrogen) = 1.01 g/mol
- S (sulfur) = 32.07 g/mol
- O (oxygen) = 16.00 g/mol
- Molar mass = 1.01 g/mol + 32.07 g/mol + (4 × 16.00 g/mol) = 98.09 g/mol
Now, let's convert the masses of KOH and H2SO4 into moles:
- Moles of KOH = mass of KOH / molar mass of KOH
= 18.00 g / 56.11 g/mol
≈ 0.321 moles of KOH (rounded to three decimal places)
- Moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
= 18.00 g / 98.09 g/mol
≈ 0.183 moles of H2SO4 (rounded to three decimal places)
Since the balanced equation tells us that 2 moles of KOH react with 1 mole of H2SO4 to form 1 mole of K2SO4, the ratio is 2:1. This means that for every 2 moles of KOH, we get 1 mole of K2SO4.
Given that we have 0.321 moles of KOH, we can calculate the moles of K2SO4 formed by using the ratio:
- Moles of K2SO4 = (0.321 moles of KOH / 2)
≈ 0.1605 moles of K2SO4 (rounded to four decimal places)
Therefore, approximately 0.1605 moles of K2SO4 should be made if all of the KOH is used.