Answer:
0.336 moles of KNO₃ are produced
The limiting reactant is Mg(NO₃)₂
Explanation:
We need to define the reaction to complete this question:
Reactants are: KOH and Mg(NO₃)₂
Products are: KNO₃ and Mg(OH)₂
The reaction is:
2KOH + Mg(NO₃)₂ → 2KNO₃ + Mg(OH)₂
We convert the mass of the reactants, to moles:
25 g / 56.1 g/mol = 0.446 moles
25 g / 148.3 g/mol = 0.168 moles
2 moles of hydroxide need 1 mol of nitrate to react
Then, 0.446 moles of KOH must need (0.446 .1) / 2 = 0.223 moles of nitrate
We do not have enough nitrate, so the Mg(NO₃)₂ is the limiting reagent.
Let's work with the equation and the ratio, which is 1:2
1 mol of Mg(NO₃)₂ produces 2 moles of KNO₃
Then, 0.168 moles of Mg(NO₃)₂ will produce (0.168 . 2) / 1 = 0.336 moles of KNO₃ are produced
