How many mL of 0.400 mol/L CaCl2 would be needed to react completely with 35.0 mL of 0.600 mol/L AgNO3 solution? The equation for the reaction is:
2AgNO3(aq) + CaCl2(aq) ------> 2AgCl(s) + Ca(NO3)2(aq)
2AgNO3(aq) + CaCl2(aq) ------> 2AgCl(s) + Ca(NO3)2(aq)
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Calculate the moles silver nitrate
35 mL = 0.035 L
0.35 x 0.6 mol/L = 0,21 moles
From the equation 2 moles silver nitrate needs 1 mole calcium chloride
so 0.21 moles silver nitrate would need 0.105 moles calcium chloride
0.105 moles / 0.400 mol/L = 0.263 L = 262 mL
Therefor you would need 262 mL
35 mL = 0.035 L
0.35 x 0.6 mol/L = 0,21 moles
From the equation 2 moles silver nitrate needs 1 mole calcium chloride
so 0.21 moles silver nitrate would need 0.105 moles calcium chloride
0.105 moles / 0.400 mol/L = 0.263 L = 262 mL
Therefor you would need 262 mL