Question

What is the mass of the precipitate formed when 50 ml of 16.9% solution of agno3 is mixed with 50ml of 5.8 nacl solution?

Answer 1

Answer:

Explanation:

50ml

Answer 2

Answer :

100 mL of AgNO3 solution contains 16.9 g of AgNO3.

Therefore, 50 mL of AgNO3 solution contains : 16.9/100*50 = 8.45 g of AgNO3.

Now, Molar mass of AgNO3 : 108 + 14 + 3*16 = 170 g/mol.

Therefore, Moles of AgNO3 present = 8.45/170 = 0.04 mol

100 mL of NaCl solution contain 5.8 g of NaCl.

Therefore, 50 mL of NaCl solution contains : 5.8/100*50 = 2.9 g of NaCl.

Now, Molar mass of NaCl : 23 + 35.5 = 58.5 g/mol.

Therefore, Moles of NaCl present = 2.9/58.5 = 0.04 mol

The reaction : AgNO3 + NaCl ➜ AgCl + NaNO3.

The precipitate formed is AgCl.

Molar mass of AgCl = 143.5 g/mol.

Amount of AgNO3 = 0.04 mol

Amount of NaCl = 0.04 mol

Hence, there is no limiting reagent.

Hence, amount of AgCl formed : 0*143.5 = 0 g.