Excess of AgNO3 was added to 2.2 g of commercial common salt dissolved in water. The mass of dried ppt. of AgCl2 was 2.11 g. Calculate the percentage purity of common salt.
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Answered by
140
The reaction is as follows:
AgNO3 + NaCl------->AgCl + NaNO3
The mole ratio is 1:1:1:1
Therefore 1mole of AgNO3 requires 1mole of NaCl to form 1 mole of AgCl.
Moles of AgCl2 formed:
Moles = mass molar mass AgCl2 (108 +35.5=143.5 )
molar mass
= 2.11/143.5 = 0.0147
mole ratio is 1:1 since moles of AgCl formed is 0.0147, the mole of Nacl used is also 0.0147.
Mass of Nacl used:
molar mass of NaCl= 58.5 (23+35.5)
mass = moles x molar mass
= 0.0147 x 58.5
= 0.85995g
percentage purity:
find the percentage 0.85995 is of 2.2
0.85995 x 100% = 39. 089 %
2.2
AgNO3 + NaCl------->AgCl + NaNO3
The mole ratio is 1:1:1:1
Therefore 1mole of AgNO3 requires 1mole of NaCl to form 1 mole of AgCl.
Moles of AgCl2 formed:
Moles = mass molar mass AgCl2 (108 +35.5=143.5 )
molar mass
= 2.11/143.5 = 0.0147
mole ratio is 1:1 since moles of AgCl formed is 0.0147, the mole of Nacl used is also 0.0147.
Mass of Nacl used:
molar mass of NaCl= 58.5 (23+35.5)
mass = moles x molar mass
= 0.0147 x 58.5
= 0.85995g
percentage purity:
find the percentage 0.85995 is of 2.2
0.85995 x 100% = 39. 089 %
2.2
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