1) Determine the quantity (g) of pure CaCl2 in 7.5 g of CaCl2•9H2O.

2. Determine the quantity (g) of pure MgSO4 in 2.4 g of MgSO4•7H2O.

3. What impact would adding twice as much Na2CO3 than required for stoichiometric quantities have on the quantity of product produced?

Answer

1)

We are given, mass of CaCl₂?9H₂O = 7.5 g and we need to calculate the mass of pure CaCl₂ in it.

For that purpose first we need to convert given mass of CaCl₂?9H₂O to its mole

We know formula for calculating moles from given mass

Moles = given mass / molar mass

Molar mass of CaCl₂?9H₂O = 273.1215 g/mole

Moles of CaCl₂?9H₂O = 7.5 g / 273.1215 g.mol^-1

                                    = 0.0275 moles CaCl₂?9H₂O

Now we know in the CaCl₂?9H₂O there is 1 CaCl₂

So,

1 moles of CaCl₂?9H₂O = 1 moles of CaCl₂

So, 0.0275 moles of CaCl₂?9H₂O = ?

= 0.0275 moles of CaCl₂

Now we calculated moles of CaCl_?2 and we know molar mass of CaCl₂

So, mass of CaC₂ = 0.0275 moles * 110.98 g/mol

                             = 3.05 g of CaCl₂

2)

We are given, mass of MgSO₄?7H₂O.= 2.4 g and we need to calculate the mass of pure MgSO₄ in it.

For that purpose first we need to convert given mass of MgSO₄?7H₂O to its mole

We know formula for calculating moles from given mass

Moles = given mass / molar mass

Molar mass of MgSO₄?7H₂O = 246.52 g/mole

Moles of MgSO₄?7H₂O = 2.4 g / 246.52 g.mol^-1

                                    = 0.00974 moles MgSO₄?7H₂O

Now we know in the MgSO₄?7H₂O there is 1 MgSO₄

So,

1 moles of MgSO₄?7H₂O= 1 moles of MgSO₄

So, 0.00974 moles of MgSO₄?7H₂O = ?

= 0.00974 moles of MgSO₄

Now we calculated moles of CaCl_?2 and we know molar mass of MgSO₄

So, mass of CaC₂ = 0.00974 moles * 120.366 g/mol

                             = 1.17 g of MgSO₄

3) When we added twice as much Na₂CO₃ than required for stoichiometric quantities then there is no any impact on the quantity of product produced, because limiting reactant are the CaCl₂ and MgSO₄ so if we increase the quantity of Na₂CO₃ then there is no effect on produced product and it is depend on the limiting reactant.