When 6.62\text{ g} of lead(II) trioxonitrate(V), \text{Pb(NO}_3)_2, is heated strongly unt… — JAMB UTME

When

6.62\text{ g}

of lead(II) trioxonitrate(V),

\text{Pb(NO}_3)_2

, is heated strongly until complete decomposition occurs according to the equation:

2\text{Pb(NO}_3)_2(s) \rightarrow 2\text{PbO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)

What is the total volume of gaseous products evolved at standard temperature and pressure (STP)?
[

M_{\text{r}}(\text{Pb(NO}_3)_2) = 331\text{ g mol}^{-1}

; Molar volume of gas at STP

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

]

The total volume of gaseous products evolved at STP is

1.12\text{ dm}^3

Decomposition of

0.02\text{ mol}

of lead(II) trioxonitrate(V) yields

0.04\text{ mol}

\text{NO}_2

and

0.01\text{ mol}

\text{O}_2

, making

0.05\text{ mol}

of total gas. At STP,

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

Calculate the amount in moles of

\text{Pb(NO}_3)_2

decomposed

\text{Moles of Pb(NO}_3)_2 = \frac{6.62\text{ g}}{331\text{ g mol}^{-1}} = 0.02\text{ mol}

Dividing given mass by relative formula mass gives the mole amount.

Determine total moles of gaseous products using stoichiometric ratios

From the balanced equation,

2\text{ mol of Pb(NO}_3)_2

yields

4\text{ mol of NO}_2(g) + 1\text{ mol of O}_2(g) = 5\text{ mol of gas}

. Total gas moles

= 0.02 \times \frac{5}{2} = 0.05\text{ mol}

Both

\text{NO}_2

and

\text{O}_2

are gases at STP, so their mole quantities must be summed.

Calculate total gas volume at STP

\text{Volume}

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

Multiplying total gaseous moles by the molar volume at STP (

22.4\text{ dm}^3\text{ mol}^{-1}

) gives the required volume.

Thermal decomposition of metallic trioxonitrate(V) salts and gas stoichiometry at STP

Estimated Time:1m 30s

Question