<u>Explanation:</u>
To calculate the number of moles, we use the equation:
......(1)
- <u>For 1:</u> 4.50 g of water
Given mass of water = 4.50 g
Molar mass of water = 18 g/mol
Putting values in equation 1, we get:
![\text{Moles of }H_2O=\frac{4.50g}{18g/mol}=0.25mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DH_2O%3D%5Cfrac%7B4.50g%7D%7B18g%2Fmol%7D%3D0.25mol)
Hence, the moles of given amount of water is 0.25 moles.
- <u>For 2:</u> 471.6 g of barium hydroxide
Given mass of barium hydroxide = 471.6 g
Molar mass of barium hydroxide = 171.34 g/mol
Putting values in equation 1, we get:
![\text{Moles of }Ba(OH)_2=\frac{471.6g}{171.34g/mol}=2.75mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DBa%28OH%29_2%3D%5Cfrac%7B471.6g%7D%7B171.34g%2Fmol%7D%3D2.75mol)
Hence, the moles of given amount of barium hydroxide is 2.75 moles.
- <u>For 3:</u> 129.68 g of iron phosphate
Given mass of iron phosphate = 129.68 g
Molar mass of iron phosphate = 150.82 g/mol
Putting values in equation 1, we get:
![\text{Moles of }Fe_3(PO_4)_2=\frac{129.68g}{150.82g/mol}=0.86mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DFe_3%28PO_4%29_2%3D%5Cfrac%7B129.68g%7D%7B150.82g%2Fmol%7D%3D0.86mol)
Hence, the moles of given amount of iron phosphate is 0.86 moles.