Answer:
No HBr will be left over by the chemical reaction.
Explanation:
The reaction of hydrobromic acid with sodium hydroxide is:
HBr + NaOH → NaBr + H₂O
80,0g of HBr are:
80,0g×(1mol/80,91g) = 0,989 moles
55,0g of NaOH are:
55,0g×(1mol/40g) = 1,375 moles
That means excess moles of NaOH are:
1,375-0,989 = 0,286 moles of NaOH
Thus, HBr is the limiting reactant and will be used completely. As a result, no HBr will be left over by the chemical reaction.
I hope it helps!
Answer:
To produce 0.9888 moles of NaBr and 0.9888 moles of H2O, there will remain 0.3862 moles of NaOH. All the HBr will be consumed, no HBr will be left over by the chemical reaction
Explanation:
Step 1: Data given
Mass of hydrobromic acid = 80.00 grams
Mass of sodium hydroxide = 55.00 grams
Molar mass HBr = 80.91 g/mol
Molar mass NaOH = 40 g/mol
Step 2: The balanced equation
HBr(aq) + NaOH(s) → NaBr(aq) + H₂O(l)
Step 3: Calculate moles of HBr
Moles HBr = Mass HBr / molar mass HBr
Moles HBr = 80.00 grams / 80.91 g/mol
Moles HBr = 0.9888 moles
Step 4: Calculate moles NaOH
Moles NaOH = 55.00 grams / 40.00 g/mol
Moles NaOH = 1.375 moles
Step 5: Calculate limiting reactant
For 1 mol of HBr we need 1 mol of NaOH to produce 1 mol of NaBr and 1 mol of H2O
The limiting reactant is HBr. It will completely be consumed. (0.9888 moles). NaOH is in excess. There will react 0.9888 moles. There will remain 1.375 - 0.9888 = 0.3862 moles
This means to produce 0.9888 moles of NaBr and 0.9888 moles of H2O, there will remain 0.3862 moles of NaOH. All the HBr will be consumed, no HBr will be left over by the chemical reaction
