If 5.3 moles of magnesium phosphate react with excess sodium chloride how many moles of magnesium chloride will be formed?

154.0831 g
Explanation:
The formula for the calculation of moles is shown below:

Given: For
Given mass = 41.0 g
Molar mass of = 24.31 g/mol
Moles of = 41.0 g / 24.31 g/mol = 1.6865 moles
Given: For
Given mass = 175 g
Molar mass of = 162.2 g/mol
Moles of = 175 g / 162.2 g/mol = 1.0789 moles
According to the given reaction:

3 moles of Mg react with 2 moles of
1 mole of Mg react with 2/3 moles of
1.6865 mole of Mg react with (2/3)*1.6865 moles of
Moles of = 1.1243 moles
Available moles of = 1.0789 moles
Limiting reagent is the one which is present in small amount. Thus, is limiting reagent. (1.0789 < 1.1243)
The formation of the product is governed by the limiting reagent. So,
2 moles of gives 3 moles of magnesium chloride
1 mole of gives 3/2 moles of magnesium chloride
1.0789 mole of gives (3/2)*1.0789 moles of magnesium chloride
Moles of magnesium chloride = 1.61835 moles
Molar mass of magnesium chloride = 95.21 g/mol
Mass of magnesium chloride = Moles × Molar mass = 1.61835 × 95.21 g = 154.0831 g