When 16.8\text{ g} of sodium hydrogentrioxocarbonate(IV) (\text{NaHCO}_3) is strongly heat… — JAMB UTME

When

16.8\text{ g}

of sodium hydrogentrioxocarbonate(IV) (

\text{NaHCO}_3

) is strongly heated in a closed system until decomposition is complete according to the equation:

2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)

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

M_r\text{ of NaHCO}_3 = 84\text{ g mol}^{-1}

; Molar volume of gas at STP

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

]

The total volume of gaseous products liberated at STP is

4.48\text{ dm}^3

Decomposing

16.8\text{ g}

(

0.20\text{ mol}

) of

\text{NaHCO}_3

yields

0.10\text{ mol}

\text{H}_2\text{O}(g)

and

0.10\text{ mol}

\text{CO}_2(g)

, totaling

0.20\text{ mol}

of gaseous products. At STP,

0.20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 4.48\text{ dm}^3

Calculate the number of moles of

\text{NaHCO}_3

decomposed.

\text{Moles of NaHCO}_3 = \frac{16.8\text{ g}}{84\text{ g mol}^{-1}} = 0.20\text{ mol}

Converting the given mass into moles using molar mass.

Determine the mole ratio between

\text{NaHCO}_3

and total gaseous products.

From

2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)

2\text{ moles}

\text{NaHCO}_3

produce

1\text{ mole}

\text{H}_2\text{O}(g)

and

1\text{ mole}

\text{CO}_2(g)

, giving

2\text{ moles}

of total gaseous products.

Both water vapor (at high decomposition temperature) and carbon(IV) oxide exist in the gaseous state.

Calculate total moles and total volume of gas at STP.

\text{Total moles of gas} = 0.20\text{ mol}

\text{Total volume} = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 4.48\text{ dm}^3

Multiplying total gaseous moles by the molar volume at STP.

Thermal decomposition stoichiometry of sodium hydrogentrioxocarbonate(IV)

Estimated Time:1m 30s

Question