Question

Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. According to the balanced equation, if 50.0mL 50.0mL of 2.0MNaCl(aq) 2.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined, the amount of precipitate formed would__________

Answer 1

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

$n=2.0 M\times 0.050 L=0.100 mol$

Moles of silver nitarte = n'

Volume of silver  nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

$NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.