<u>Answer:</u> The mass of NaBr that can be produced is 6.3 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of HBr = 17 g
Molar mass of HBr = 81 g/mol
Putting values in equation 1, we get:
![\text{Moles of HBr}=\frac{17g}{81g/mol}=0.210mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20HBr%7D%3D%5Cfrac%7B17g%7D%7B81g%2Fmol%7D%3D0.210mol)
Given mass of NaOH = 2.44 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
![\text{Moles of NaOH}=\frac{2.44g}{40g/mol}=0.061mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20NaOH%7D%3D%5Cfrac%7B2.44g%7D%7B40g%2Fmol%7D%3D0.061mol)
The chemical equation for the reaction of HBr and NaOH follows:
![HBr+NaOH\rightarrow NaBr+H_2O](https://tex.z-dn.net/?f=HBr%2BNaOH%5Crightarrow%20NaBr%2BH_2O)
By Stoichiometry of the reaction:
1 mole of NaOH reacts with 1 mole of HBr
So, 0.061 moles of NaOH will react with =
of HBr
As, given amount of HBr is more than the required amount. So, it is considered as an excess reagent.
Thus, NaOH is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of NaOH produces 1 mole of NaBr
So, 0.061 moles of NaOH will produce =
of carbon dioxide
Now, calculating the mass of NaBr from equation 1, we get:
Molar mass of NaBr = 103 g/mol
Moles of NaBr = 0.061 moles
Putting values in equation 1, we get:
![0.061mol=\frac{\text{Mass of NaBr}}{103g/mol}\\\\\text{Mass of NaBr}=(0.061mol\times 103g/mol)=6.28g](https://tex.z-dn.net/?f=0.061mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20NaBr%7D%7D%7B103g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20NaBr%7D%3D%280.061mol%5Ctimes%20103g%2Fmol%29%3D6.28g)
Hence, the mass of NaBr that can be produced is 6.3 grams