Question

if 25 g of potassium nitrate hydroxide reacts with 25bg of magneisum nitartae to produce potassium nitrate and magneisum hydroxide. Determine how much potassium nitrate is produced. what is the limiting reactsnt

Answer 1

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