Answer:
0.35 g of H₂O
Explanation:
Firstly we have to make the reaction's equation.
In this case, the reactants are HBr and NaOH and the products are water and NaBr. This is a neutralization reaction.
We calculate the moles of each reactant:
1.6 g / 80.9 g/mol = 0.0198 moles of HBr
1.15 g / 39 g/mol = 0.0295 moles of NaOH
Reaction: NaOH + HBr → NaBr + H₂O
Ratio is 1:1, so the limiting reactant is the hydrobromic acid.
For 0.0295 moles of hydroxide we need the same amount of acid, but we do not have enough HBr, just 0.0198 moles.
Ratio is 1:1, again. 1 mol of acid can produce 1 mol of water.
Therefore, 0.0198 moles of HBr will produce 0.0198 moles of water.
We convert the moles to mass → 0.0198 mol . 18 g/mol = 0.35 g of H₂O
Answer:
0.36 grams of H2O will be produced
Explanation:
Step 1: data given
Mass of hydrobromic acid (HBr) = 1.6 grams
Molar mass HBr = 80.91 g/mol
Mass of sodium hydroxide (NaOH) = 1.15 grams
Molar mass of NaOH = 40.0 g/mol
Step 2: The balanced equation
HBr + NaOH → NaBr + H2O
Step 3: Calculate moles HBr
Moles HBr = mass HBr / molar mass HBr
Moles HBr = 1.6 grams / 80.91 g/mol
Moles HBr = 0.0198 moles
Step 4: Calculate moles NaOH
Moles NaOH = 1.15 grams / 40.0 g/mol
Moles NaOH = 0.0288 moles
Step 5: Calculate limiting reactant
For 1 mol HBr we need 1 mol NaOH to produce 1 mol NaBr and 1 mol H2O
HBr is the limiting reactant. It will completely be consumed (0.0198 moles). NaOH is in excess. There will react 0.019 moles. There will remain 0.0288 - 0.0198 = 0.009 moles
Step 6: Calculate moles H2O
For 1 mol HBr we need 1 mol NaOH to produce 1 mol NaBr and 1 mol H2O
For 0.0198 moles HBr we'll have 0.0198 moles of H2O
Step 7: Calculate mass H2O
Mass H2O = 0.0198 moles * 18.02 g/mol
Mass H2O = 0.36 grams
0.36 grams of H2O will be produced
