Answer:
11.25 g of water, is the maximum mass that can be produced by the reaction
Explanation:
We propose the reaction, where aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water
The equation is: HBr (aq) + NaOH(s) → NaBr(aq) + H₂O(l)
Ratio is 1:1.
We convert the mass of the reactants to moles, in order to find out the limiting reagent
68 g / 80.90 g/mol = 0.840 moles of HBr
25 g / 40 g/mol = 0.625 moles of NaOH
Limiting reagent is the NaOH. For 0.840 moles of HBr, we need 0.840 moles of NaOH, but we only have 0.625.
To calculate the mass of water that could be produced, the ratio is 1:1
Then, 0.625 moles of NaOH will produce 0.625 moles of water.
We convert the moles to mass, to determine the maximum mass that can be produced → 0.625 mol . 18 g / 1mol = 11.25 g
Answer:
11.3 grams of water will be produced.
Explanation:
Step 1: Data given
Mass of hydrobromic acid (HBr) = 68.0 grams
Molar mass of HBr = 80.91 g/mol
Mass of sodium hydroxide ( NaOH) = 25.0 grams
Molar mass of NaOH = 40.0 g/mol
Step 2: The balanced equation
HBr + NaOH → NaBr + H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles HBr = 68.0 grams / 80.91 g/mol
Moles HBr = 0.840moles
Moles NaOH = 25.0 grams / 40.0 g/mol
Moles NaOH = 0.625 moles
Step 4: Calculate the limiting reactant
NaOH is the limiting reactant. It will completely be consumed (0.625 moles).
HBr is in excess. There will react 0.625 moles. There will remain 0.840 - 0.625 = 0.215 moles NaOH
Step 5: Calculate moles H2O
For 1 mol HBr we need 1 mol NaOH to produce 1 mol NaBr and 1 mol H2O
For 0.625 moles NaOH we'll have 0.625 moles H2O
Step 6: Calculate mass of H2O
Mass of H2O = moles H2O * molar mass H2O
Mass H2O = 0.625 moles * 18.02 g/mol
Mass H2O = 11.3 grams H2O
11.3 grams of water will be produced.
