Answer:
6.2g of NaBr are produced
Step-by-step explanation:
The reaction of HBr with NaOH occurs as follows:
HBr + NaOH → NaBr + H2O
Where 1 mole of each reactant produce 1 mole of NaBr
To solve this question we need to find the moles of each reactant using their molar mass. With moles we can find limiting reactant and the moles (And mass) of NaBr produced, as follows:
Moles HBr -Molar mass: 80.9119g/mol)-
4.9g * (1mol/80.9119g) = 0.0606 moles HBr
Moles NaOH -Molar mass: 40g/mol-
3.86g * (1mol/40g) = 0.0965 moles NaOH
As the reaction is 1:1 and the moles of HBr < Moles NaOH, the limiting reactant is HBr and moles of NaBr produced are 0.0606 moles.
The mass of NaBr (Molar mass: 102.894g/mol) is:
0.0606 moles * (102.894g/mol) =
6.2g of NaBr are produced