Respuesta :
Answer:
1.822 g of magnesium hydroxide would be produced.
Explanation:
Balanced reaction: [tex]2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl[/tex]
Compound Molar mass (g/mol)
NaOH 39.997
[tex]MgCl_{2}[/tex] 95.211
[tex]Mg(OH)_{2}[/tex] 58.3197
So, 2.50 g of NaOH = [tex]\frac{2.50}{39.997}[/tex] mol of NaOH = 0.0625 mol of NaOH
4.30 g of [tex]MgCl_{2}[/tex] = [tex]\frac{4.30}{95.211}[/tex] mol of [tex]MgCl_{2}[/tex] = 0.0452 mol of [tex]MgCl_{2}[/tex]
According to balanced equation-
2 mol of NaOH produce 1 mol of [tex]Mg(OH)_{2}[/tex]
So, 0.0625 mol of NaOH produce [tex](\frac{0.0625}{2})[/tex] mol of NaOH or 0.03125 mol of NaOH
1 mol of [tex]MgCl_{2}[/tex] produces 1 mol of [tex]Mg(OH)_{2}[/tex]
So, 0.0452 mol of [tex]MgCl_{2}[/tex] produce 0.0452 mol of [tex]Mg(OH)_{2}[/tex]
As least number of moles of [tex]Mg(OH)_{2}[/tex] are produced from NaOH therefore NaOH is the limiting reagent.
So, amount of [tex]Mg(OH)_{2}[/tex] would be produced = 0.03125 mol
= [tex](0.03125\times 58.3197)[/tex] g
= 1.822 g
