Answer:
To solve this problem, we first need to determine the molar mass of Na2CO3 (sodium carbonate) and then use stoichiometry to find out how many grams of NaOH (sodium hydroxide) are needed to produce 204.68 g of Na2CO3.
1. **Molar mass of Na2CO3**:
Na: 22.99 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of Na2CO3 = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 2(22.99 g/mol) + 12.01 g/mol + 48.00 g/mol = 46.00 g/mol + 12.01 g/mol + 48.00 g/mol = 106.01 g/mol
2. **Stoichiometry**:
From the balanced equation, 2 moles of NaOH react with 1 mole of Na2CO3.
3. **Calculate moles of Na2CO3**:
Moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
= 204.68 g / 106.01 g/mol
≈ 1.932 moles
4. **Calculate moles of NaOH**:
According to the stoichiometry, 2 moles of NaOH are needed for 1 mole of Na2CO3.
Moles of NaOH = (2/1) * Moles of Na2CO3
= 2 * 1.932 moles
≈ 3.864 moles
5. **Calculate grams of NaOH**:
Mass of NaOH = Moles of NaOH * Molar mass of NaOH
= 3.864 moles * 40.00 g/mol
≈ 154.56 grams
So, approximately 154.56 grams of NaOH are needed to produce 204.68 grams of Na2CO3.
