Consider the following chemical reaction:
NaOH +HC1 -> NaC1 + H20
When 36.5 g of HC1 are reacted with 40.0 g of NaOH , how many grams of water are produced along with 58.5 g of NaC1?
..........NaOH + HCl ==> NaCl + H2O
moles NaOH = grams/molar mass = 40/40 = 1.o mole.
moles HCl = 36.5/36.5 = 1.0 mole
So 1 mole reacts with 1 mol to produce 1 mole of NaCl(58.5g) and 1 mole H2O.
grams H2O = mole x molar mass = ??
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To determine the amount of water produced in the chemical reaction, we need to use the concept of stoichiometry. Stoichiometry is the calculation of quantities in chemical reactions based on the balanced equation.
1. Start by writing the balanced equation for the reaction:
NaOH + HC1 -> NaC1 + H2O
2. Identify the molar masses of the substances involved:
- NaOH: 22.99 g/mol (Sodium - 22.99 g/mol + Oxygen - 16.00 g/mol + Hydrogen - 1.01 g/mol)
- HC1: 36.46 g/mol (Hydrogen - 1.01 g/mol + Chlorine - 35.45 g/mol)
- NaC1: 58.44 g/mol (Sodium - 22.99 g/mol + Chlorine - 35.45 g/mol)
- H2O: 18.02 g/mol (Hydrogen - 1.01 g/mol + Oxygen - 16.00 g/mol)
3. Convert the given masses of NaOH, HC1, and NaC1 into moles using the molar masses:
- Moles of NaOH = 40.0 g NaOH / 22.99 g/mol NaOH = 1.74 mol NaOH
- Moles of HC1 = 36.5 g HC1 / 36.46 g/mol HC1 = 1.00 mol HC1
- Moles of NaC1 = 58.5 g NaC1 / 58.44 g/mol NaC1 = 1.00 mol NaC1
4. Use the balanced equation to determine the mole ratio between HC1 and H2O:
From the balanced equation: 1 mol HC1 : 1 mol H2O
5. Since the mole ratio is 1:1, the number of moles of water produced is equal to the number of moles of HC1 that reacted:
Moles of H2O = 1.00 mol HC1
6. Convert the moles of H2O into grams using the molar mass of H2O:
Grams of H2O = 1.00 mol H2O x 18.02 g/mol H2O = 18.02 g H2O
Therefore, along with 58.5 g of NaCl, the reaction produces 18.02 g of water.