A 7.56 g mixture of Pb(NO3)2 and NaNO3…

A 7.56 g mixture of Pb(NO3)2 and NaNO3 is heated strongly until decomposition is complete. The only gaseous products are NO2 and O2 (treat NO2 as not…

Single correct option

7.56\ \mathrm{g}

mixture of

\mathrm{Pb(NO_3)_2}

and

\mathrm{NaNO_3}

is heated strongly until decomposition is complete. The only gaseous products are

\mathrm{NO_2}

and

\mathrm{O_2}

(treat

\mathrm{NO_2}

as not dimerising), and their total volume is

1.12\ \mathrm{L}

273\ \mathrm{K}

and

1\ \mathrm{atm}

(molar volume

22.4\ \mathrm{L}

). The solid residue consists of

\mathrm{PbO}

and

\mathrm{NaNO_2}

. Taking molar masses

\mathrm{Pb(NO_3)_2}=331.2

\mathrm{NaNO_3}=85.0

\mathrm{PbO}=223.2

\mathrm{NaNO_2}=69.0\ \mathrm{g\,mol^{-1}}

, the mass of the solid residue is

A. $5.96\ \mathrm{g}$
B. $5.68\ \mathrm{g}$
C. $6.40\ \mathrm{g}$
D. $3.78\ \mathrm{g}$

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Answer: (B) $5.68\ \mathrm{g}$

The decompositions are

\mathrm{Pb(NO_3)_2\to PbO+2NO_2+\tfrac12 O_2}

, giving

2.5

mol of gas per mole, and

\mathrm{NaNO_3\to NaNO_2+\tfrac12 O_2}

, giving

0.5

mol of gas per mole. Total gas is

1.12/22.4=0.0500

mol. Let

x

and

y

be the moles of the two nitrates. Then

331.2x+85.0y=7.56

and

2.5x+0.5y=0.0500

. Solving gives

x=0.0100

and

y=0.0499\approx0.0500

mol. The residue is

x(223.2)+y(69.0)=2.232+3.45=5.68\ \mathrm{g}

. As a check, the mass lost as gas is

0.0100(2\times46+16)+0.0500(16)=1.08+0.80=1.88\ \mathrm{g}

, and

7.56-1.88=5.68\ \mathrm{g}

. The value

3.78\ \mathrm{g}

comes from wrongly assuming

\mathrm{NaNO_3}

gives

\mathrm{Na_2O}

, which ignores the nitrite product stated in the question.

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Original question written in the JEE Advanced 2017 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake