The molarity of the Ca(NO3)2 solution would be 0.005 M
First, let us look at the balanced equation of the reaction:
[tex]2HNO_3+Ca(OH)_2 ---> Ca(NO_3)_2+2H_2O[/tex]
The mole ratio of HNO3 to Ca(OH)2 is 2:1.
Mole of HNO3 = molarity x volume
= 0.2 x 150/1000
= 0.03 mole
Mole of Ca(OH)2 = 0.01 x 150/1000
= 0.0015
Thus, there is no limiting or excess reactant.
Also from the equation, mole ratio of Ca(OH)2 to Ca(NO3)2 is 1:1. Hence, the mole of Ca(NO3)2 would also be 0.0015 mole.
The total volume of the resulting solution would be: 150 +150 = 300 mL
Thus, the molarity of the resulting Ca(NO3)2 would be:
Molarity = mole/volume
= 0.0015/0.3
= 0.005 M
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