When 6.62\text{ g} of lead(II) trioxonitrate(V) is completely decomposed by heating accord… — JAMB UTME

When

6.62\text{ g}

of lead(II) trioxonitrate(V) is completely decomposed by heating according to the balanced chemical equation:

2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)

What is the total volume of gaseous products liberated at STP?
[

1\text{ mole of gas at STP} = 22.4\text{ dm}^3

; relative atomic masses:

Pb = 207

N = 14

O = 16

]

1.12\text{ dm}^3

The correct answer is

1.12\text{ dm}^3

. Decomposing

6.62\text{ g}

(

0.02\text{ mol}

) of

Pb(NO_3)_2

yields

0.04\text{ mol}

NO_2

and

0.01\text{ mol}

O_2

, totaling

0.05\text{ mol}

of gas. At STP,

0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

Calculate the molar mass of lead(II) trioxonitrate(V),

Pb(NO_3)_2

\text{Molar Mass} = 207 + 2 \times (14 + 3 \times 16) = 331\text{ g mol}^{-1}

Molar mass is needed to convert the given mass into moles of reactant.

Determine the amount (in moles) of

Pb(NO_3)_2

reacted.

n(Pb(NO_3)_2) = \frac{6.62\text{ g}}{331\text{ g mol}^{-1}} = 0.02\text{ mol}

Quantitative stoichiometric relations require knowing the exact mole quantity of the reactant.

Identify the total mole ratio of gaseous products to reactant from the balanced chemical equation.

2\text{ moles } Pb(NO_3)_2 \rightarrow 4\text{ moles } NO_2(g) + 1\text{ mole } O_2(g) = 5\text{ moles of total gas}

. Total gas mole ratio

= \frac{5}{2} = 2.5

Both

NO_2

and

O_2

are gases at STP, so both contribute to the total volume evolved.

Calculate the total moles and volume of gas liberated at STP.

\text{Total moles of gas} = 0.02 \times 2.5 = 0.05\text{ mol}

\text{Total volume at STP} = 0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

Multiplying total gaseous moles by the standard molar gas volume gives the total volume at STP.

Mass-Volume Stoichiometric Calculation for Reaction Systems Yielding Multiple Gaseous Products

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