The mass of the precipitate is 8.74 g and the concentration of the calcium nitrate is 0.061 M.
What mass of precipitate is formed?
The reaction is given by; 2AgNO3(aq) + CaCl2 (aq) ------>2AgCl(s) + Ca(NO3)2(aq)
Number of moles of AgNO3 = 195/1000 * 0.500 = 0.0975 moles
Number of moles of CaCl2 = 305/1000 * 0.100 = 0.0305 moles
If 2 moles of AgNO3 reacts with 1 mole of CaCl2
0.0975 moles reacts with 0.0975 moles * 1 mole / 2 moles = 0.04875 moles
Thus CaCl2 is the limiting reactant.
1 mole of CaCl2 produces 2moles of AgCl
0.0305 moles produces 0.0305 moles * 2 moles/ 1mole
= 0.061 moles
Mass of precipitate = 0.061 moles * 143.32 g/mol
= 8.74 g
Now 1 mole of CaCl2 produces 1 mole of Ca(NO3)2
Total volume of solution = 195 ml + 305 ml = 500 ml or 0.5 L
Concentration of Ca(NO3)2 = 0.0305 moles/ 0.5 L
= 0.061 M
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