The mass of sodium nitrate and calcium chloride produced would be 66.3 g and 43.3 g respectively.
The equation of the reaction between sodium chloride and calcium nitrate is given by the following equation:
2NaCl + Ca(NO3)2 ---> CaCl2 + 2NaNO3
Mole ratio of NaCl and Ca(NO3)2 = 2:1
Mole of 45.8g NaCl = 45.8/58.44
= 0.78 moles
Mole of 75.0 g Ca(NO3)2 = 75/164.1
= 0.46 moles
Equivalent mole of Ca(NO3)2 = 0.78/2 = 0.39
Ca(NO3)2 is in excess. Thus NaCl is the limiting reactant.
Mole ratio of NaCl and CaCl2 = 2;1
Equivalent mole of CaCl2 = 0.78/2 = 0.39
Mass of 0.39 CaCl2 = 0.39 x 110.98
= 43.3 g
Mole ratio of NaCl and NaNO3 = 1:1
Equivalent mole of NaNO3 = 0.78/1 = 0.78
Mass of 0.78 mole NaNO3 = 0.78 x 84.99
= 66.3 g
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