Sodium hydrogentrioxocarbonate(IV) decomposes upon heating according to the balanced chemi… — JAMB UTME

Sodium hydrogentrioxocarbonate(IV) decomposes upon heating according to the balanced chemical equation:

2NaHCO_3(s) \rightarrow Na_2CO_3(s) + H_2O(l) + CO_2(g)

What mass of sodium trioxocarbonate(IV) (

Na_2CO_3

) is produced by the complete decomposition of

16.8\text{ g}

NaHCO_3

[Na = 23, C = 12, O = 16, H = 1]

The mass of sodium trioxocarbonate(IV) produced is

10.6\text{ g}

The complete decomposition of

16.8\text{ g}

(

0.20\text{ mol}

) of

NaHCO_3

produces

0.10\text{ mol}

Na_2CO_3

according to the

2:1

mole ratio in the balanced equation. Multiplying

0.10\text{ mol}

by the molar mass of

Na_2CO_3

(

106\text{ g mol}^{-1}

) gives

10.6\text{ g}

Calculate the molar masses of

NaHCO_3

and

Na_2CO_3

Molar mass of

NaHCO_3 = 23 + 1 + 12 + 3(16) = 84\text{ g mol}^{-1}

. Molar mass of

Na_2CO_3 = 2(23) + 12 + 3(16) = 106\text{ g mol}^{-1}

Molar masses are required to convert between mass and mole amounts.

Determine the number of moles of

NaHCO_3

reacted

\text{Moles of } NaHCO_3 = \frac{16.8\text{ g}}{84\text{ g mol}^{-1}} = 0.20\text{ mol}.

Stoichiometric relations are calculated using mole amounts rather than raw masses.

Use the mole ratio from the balanced chemical equation to find moles of

Na_2CO_3

From

2NaHCO_3 \rightarrow Na_2CO_3

, ratio is

2:1

. Therefore, \text{moles of } Na_2CO_3 = \frac{0.20}{2} = 0.10\text{ mol}.

Two moles of

NaHCO_3

produce one mole of

Na_2CO_3

Calculate the mass of

Na_2CO_3

produced

\text{Mass of } Na_2CO_3 = 0.10\text{ mol} \times 106\text{ g mol}^{-1} = 10.6\text{ g}.

Multiplying moles of product by its molar mass yields the theoretical yield mass.

Mass-Mass Stoichiometric Calculations

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