Question

A sample of naclo3 is converted by heat to nacl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as agcl. The mass of agcl (in g) obtained will be : (given : molar mass of agcl=143.5 g molâ’1)

Answer 1

Answer:- 0.47 gram AgCl.

Solution:- The balanced equation for the decomposition of sodium chlorate is written as below:

$2NaClO_3\rightarrow 2NaCl+3O_2$

Using the given mass of oxygen and the mol ratio the moles of NaCl are calculated as:-

$0.16gO_2(\frac{1molO_2}{32gO_2})(\frac{2molNaCl}{3molO_2})$

= 0.0033 moles NaCl

NaCl is dissolved in water to make its aqueous solution to which some silver compound solution like silver nitrate is added to get the precipitate of AgCl.

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

There is 1:1 mol ratio between NaCl and AgCl, so moles of AgCl formed will also be 0.0033 that are caculated as:

$0.0033molesNaCl(\frac{1molAgCl}{1molNaCl})$

= 0.0033 moles AgCl

Moles of AgCl are converted to grams on multiplying by molar mass.

$0.0033molAgCl(\frac{143.5gAgCl}{1molAgCl})$

= 0.47 g AgCl

So, the mass of AgCl obtained will be 0.47 grams.

Answer 2

Hey dear,

● Answer -
W(AgCl) = 0.4736 g

● Explaination-
Initially,
2NaClO3 ---> 2NaCl + 3O2
Hence,
n(NaCl)/2 = n(O2)/3 ...(1)

Here,
n(O2) = 0.16/32 = 0.005 mol

Putting this in eqn (1),
n(NaCl)/2 = 0.005/3
n(NaCl) = 0.0033 mol

Later,
NaCl + Ag ---> AgCl + Na
Hence,
n(NaCl) = n(AgCl)
n(AgCl) = 0.0033 mol

Mass of AgCl precipitated,
W(AgCl) = 0.0033 × 143.5
W(AgCl) = 0.4736 g

Hope this is useful...