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

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

the answers would be

the first one and the second one

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8