Answer : 29.475 grams of water can be produced.
Solution : Given,
Mass of NaOH = 65.5 g
Molar mass of NaOH = 40 g/mole
Molar mass of
= 18 g/mole
The given balanced equation is,
![H_2SO_4+2NaOH\rightarrow 2H_2O+Na_2SO_4](https://tex.z-dn.net/?f=H_2SO_4%2B2NaOH%5Crightarrow%202H_2O%2BNa_2SO_4)
First we have to calculate the moles of NaOH.
![\text{ Moles of NaOH}=\frac{\text{ Mass of NaOH}}{\text{ Molar mass of NaOH}}=\frac{65.5g}{40g/mole}=1.6375moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20NaOH%7D%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20NaOH%7D%7D%7B%5Ctext%7B%20Molar%20mass%20of%20NaOH%7D%7D%3D%5Cfrac%7B65.5g%7D%7B40g%2Fmole%7D%3D1.6375moles)
Now we have to calculate the moles of
.
From the given reaction, we conclude that
As, 2 moles of NaOH react to give 2 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
So, 1 mole of NaOH react to give 1 mole of ![H_2O](https://tex.z-dn.net/?f=H_2O)
And, 1.6375 moles of NaOH react to give 1.6375 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
The moles of
= 1.6375 moles
Now we have to calculate the mass of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Mass of
= Moles of
× Molar mass of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Mass of
= 1.6375 moles × 18 g/mole = 29.475 g
Therefore, 29.475 grams of water can be produced.