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