Question #76947

Can we change limiting reactant in chemical reaction to increase the amount of required product?
Like if zinc nitrate hexahydrate (15.0 g) reacts with sodium hydroxide (16.0 g) to produce zinc oxide (which was just 4.0 g), then can we change the amount of reactants in such a way so that the amount of zinc oxide produced is more?

Expert's answer

Answer on Question #76947 – Chemistry – General Chemistry

Can we change limiting reactant in chemical reaction to increase the amount of required product? Like if zinc nitrate hexahydrate (15.0 g) reacts with sodium hydroxide (16.0 g) to produce zinc oxide (which was just 4.0 g), then can we change the amount of reactants in such a way so that the amount of zinc oxide produced is more?

Solution:


Zn(NO3)2+2NaOHZn(OH)2+2NaNO3Zn(OH)2ZnO+H2On(Zn(NO3)26H2O)=m/M=15.0 g/297.0 g/mol=0.05 moln(NaOH)=16.0 g/40.0 g/mol=0.4 moln(ZnO)=4.0 g/81.0 g/mol=0.05 mol\begin{array}{l} \mathrm{Zn(NO_3)_2 + 2NaOH \rightarrow Zn(OH)_2 + 2NaNO_3} \\ \mathrm{Zn(OH)_2 \rightarrow ZnO + H_2O} \\ \mathrm{n(Zn(NO_3)_2 \cdot 6H_2O) = m/M = 15.0\ g / 297.0\ g/mol = 0.05\ mol} \\ \mathrm{n(NaOH) = 16.0\ g / 40.0\ g/mol = 0.4\ mol} \\ \mathrm{n(ZnO) = 4.0\ g / 81.0\ g/mol = 0.05\ mol} \\ \end{array}


Zinc nitrate hexahydrate is limiting reactant

If we want more zinc oxide to be produced we must increase amount of Zn(NO3)26H2O\mathrm{Zn(NO_3)_2 \cdot 6H_2O}

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