Question

Объясните пожалуйста какую соль правильнее выбрать из следующих двух реакции если избыток 2,43 (так как количество KOH - 0,78 моль и количество H2SO4 - 0,32 моль)??????????????????? 1) H2SO4+2KOH →K2SO4+2H2O. 2) H2SO4+KOH→KHSO4+H2O.

Answer 1

Воспользуемся правилом: если количество щелочи больше, чем кислоты, то образуется СРЕДНЯЯ соль, если же кислоты больше, то КИСЛАЯ соль образуется.
В данном задании количество щелочи больше, следовательно образуется средняя соль:
H2SO4+2KOH →K2SO4+2H2O.

Answer 2

Choosing Between Two Reactions for Excess 2.43 Moles of Reactants

When deciding which reaction to choose given an excess of 2.43 moles of reactants (0.78 moles of KOH and 0.32 moles of H2SO4), it's important to consider the balanced chemical equations for the reactions and the stoichiometry of the reactants and products.

The two reactions to consider are: 1. H2SO4 + 2KOH → K2SO4 + 2H2O 2. H2SO4 + KOH → KHSO4 + H2O

To determine which reaction is more suitable, we need to calculate the limiting reactant and the amount of excess reactant in each case.

First, let's calculate the moles of each reactant and identify the limiting reactant.

The balanced chemical equations are: 1. H2SO4 + 2KOH → K2SO4 + 2H2O 2. H2SO4 + KOH → KHSO4 + H2O

Given: - Moles of KOH = 0.78 - Moles of H2SO4 = 0.32

Calculating the Limiting Reactant

Using the stoichiometry of the reactions, we can calculate the moles of products that would be formed from the given moles of reactants.

For Reaction 1: - Moles of K2SO4 produced = (0.32 moles of H2SO4) / 1 * (2 moles of KOH) / 1 = 0.64 moles - Moles of H2O produced = (0.32 moles of H2SO4) / 1 * (2 moles of KOH) / 1 * 2 = 1.28 moles

For Reaction 2: - Moles of KHSO4 produced = 0.32 moles of H2SO4 - Moles of H2O produced = 0.32 moles of H2SO4

Identifying the Excess Reactant

The excess reactant is the reactant that is not completely consumed in the reaction. To find the excess reactant, we compare the moles of each reactant with the moles required by the reaction.

For Reaction 1: - Excess KOH = 0.78 moles - 0.64 moles = 0.14 moles - Excess H2SO4 = 0.32 moles - 0.32 moles = 0 moles

For Reaction 2: - Excess KOH = 0.78 moles - 0.32 moles = 0.46 moles - Excess H2SO4 = 0.32 moles - 0.32 moles = 0 moles

Conclusion

Based on the calculations, it is clear that Reaction 1: H2SO4 + 2KOH → K2SO4 + 2H2O is the more suitable reaction when there is an excess of 2.43 moles of reactants. This reaction results in less excess reactant and a higher yield of products compared to Reaction 2.

Therefore, the correct reaction to choose when there is an excess of 2.43 moles of reactants is Reaction 1: H2SO4 + 2KOH → K2SO4 + 2H2O.