If we combined 6 mol KOH and 6 mol H2SO4, and they reacted via the balanced equation below, how many moles of K2SO4 should be made?
2 KOH + H2SO4 = 2 H2O + K2SO4
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If there are no instructions about which reacts completely, this means you have a limiting regent (LR) problem. You know that is the case when an amount is given for more than one of the reactants. So you look at the equation and you know 6 mols H2SO4 will produce 6 moles K2SO4 because 1 mol H2SO4 produces 1 mol K2SO4. Likewise, 6 mols KOH will produce just 3 mols K2SO4 so KOH is the LR. So you get 3 mols K2SO4 and you will have some H2SO4 left over unreacted.
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To determine how many moles of K2SO4 will be produced when 6 mol of KOH and 6 mol of H2SO4 react, we need to use the balanced equation provided and the concept of stoichiometry.
The balanced equation is:
2 KOH + H2SO4 --> 2 H2O + K2SO4
From the balanced equation, we can see that for every 2 moles of KOH, we obtain 1 mole of K2SO4. Therefore, we need to find the ratio of KOH to K2SO4 in order to determine the number of moles of K2SO4 produced.
The ratio of KOH to K2SO4 is:
2 moles KOH : 1 mole K2SO4
Since we have 6 moles of KOH, we can set up a ratio to calculate the moles of K2SO4:
6 mol KOH * (1 mol K2SO4 / 2 mol KOH) = 3 mol K2SO4
Therefore, 6 moles of KOH reacting with 6 moles of H2SO4 will produce 3 moles of K2SO4.