Respuesta :
Answer:
NaOH is the limiting reactant and MgCl2 is in excess.
Explanation:
let's calculate the mole of each reactant.
number of moles of MgCl2 = [tex]\frac{187}{1000} L* 0.103 mol / L =0.01926 mol\\[/tex]
number of moles of NaOH = [tex]\frac{33.58 }{1000}L * 0.839 mol/L =0.0282 mol[/tex]
as per the reaction , 1 mol of Magnesium choloride reacts with 2 moles of NaOH
hence 0.01926 needs 0.01926x2= 0.03852 moles of NaOH , but there are only 0.0282 moles of NaOH available .
The limiting reactant for the reaction between 187.0 mL solution of 0.103 M MgClâ‚‚ and 33.58 mL solution of 0.839 M NaOH is NaOH
We'll begin by calculating the number of mole of MgClâ‚‚ and NaOH present in the solution.
For MgClâ‚‚:
Volume = 187 mL = 187 / 1000 = 0.187 L
Molarity = 0.103 M
Mole of MgClâ‚‚ =?
Mole = Molarity x Volume
Mole of MgCl₂ = 0.103 × 0.187
Mole of MgClâ‚‚ = 0.019 mole
For NaOH:
Volume = 33.58 mL = 33.58 / 1000 = 0.03358 L
Molarity = 0.839 M
Mole of NaOH =?
Mole = Molarity x Volume
Mole of NaOH = 0.839 × 0.03358
Mole of NaOH = 0.028 mole
Finally, we shall determine the limiting reactant. This can be obtained as follow:
MgCl₂ + 2NaOH —> Mg(OH)₂ + 2NaCl
From the balanced equation above,
2 moles of NaOH required 1 mole of MgClâ‚‚.
Therefore, 0.028 mole of NaOH will require = [tex]\frac{0.0282}{2}\\[/tex] = 0.014 mole of MgClâ‚‚
From the calculations made above, we can see that only 0.014 mole of MgClâ‚‚ out of 0.019 mole reacted completely with 0.028 mole of NaOH.
Therefore, NaOH is the limiting reactant and MgClâ‚‚ is the excess reactant.
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