Answer:
0.04mol of NaCl and 0.04mol of AgNO3
Explanation:
To calculate the numbers of mole of NaCl and AgNO3 used in the reaction, first we obtained their mole before the reaction as illustrated below:
For NaCl:
Molarity = 0.2M
Volume = 200mL = 200/1000 = 0.2L
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.2 x 0.2 = 0.04mol
For AgNO3:
Molarity = 0.5M
Volume = 100mL = 100/1000 = 0.1L
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.5 x 0.1 = 0.05mol
NaCl + AgNO3 —> AgCl + NaNO3
From the above equation,
1mole of NaCl required 1mole of AgNO3.
Therefore for, 0.04mol of NaCl will also require 0.04mol of AgNO3. This implies that AgNO3 is the excess reactant
