A sample of 0.13\text{ g} of zinc granules reacts completely with an excess of dilute hydr… — JAMB UTME

A sample of

0.13\text{ g}

of zinc granules reacts completely with an excess of dilute hydrochloric acid according to the reaction equation:

\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)

If the reaction takes exactly

40\text{ seconds}

to reach completion, what is the average rate of consumption of hydrochloric acid in

\text{mol s}^{-1}

? (Molar mass of

\text{Zn} = 65\text{ g mol}^{-1}

)

Cevap: 0.0001 mol s^-1

The average rate of consumption of hydrochloric acid is

0.0001\text{ mol s}^{-1}

(or

1.0 \times 10^{-4}\text{ mol s}^{-1}

To determine the average rate of consumption of hydrochloric acid, first convert the mass of zinc to moles (

0.13\text{ g} / 65\text{ g mol}^{-1} = 0.002\text{ mol}

). According to the stoichiometric coefficients in the balanced equation

\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2

2\text{ moles}

\text{HCl}

react for every

1\text{ mole}

\text{Zn}

. Therefore,

0.004\text{ mol}

\text{HCl}

is consumed. Dividing this quantity by the reaction time (

40\text{ seconds}

) gives an average rate of

0.0001\text{ mol s}^{-1}

Calculate the amount in moles of zinc reacted

\text{Moles of Zn} = \frac{0.13\text{ g}}{65\text{ g mol}^{-1}} = 0.002\text{ mol}

Mass divided by molar mass yields the quantity in moles.

Determine the moles of hydrochloric acid consumed using the mole ratio

\text{Moles of HCl} = 2 \times 0.002\text{ mol} = 0.004\text{ mol}

The balanced chemical equation shows a

1:2

stoichiometric ratio between

\text{Zn}

and

\text{HCl}

Calculate the average rate of consumption of HCl per unit time

\text{Rate of HCl consumption} = \frac{0.004\text{ mol}}{40\text{ s}} = 0.0001\text{ mol s}^{-1}

Rate of reaction is defined as the change in moles of reactant divided by elapsed time.

Stoichiometric determination of reaction rate from reactant consumption

Soru