First write the balanced reaction:
2 NaOH + MgCl2 ---> Mg(OH)2 + 2 NaCl
Then we determine the moles of each reactant using the given volume and concentration (molarity). Recall that molarity is moles of solute per liter solution, or
M = n / V
Solving for moles,
n = M * V
MgCl2:
n = 0.325 M * 0.148 L = 0.0481 mol MgCl2
NaOH:
n = 0.388 M * 0.0687 L = 0.02665 mol NaOH
We don't know yet which is the limiting reactant, but we can determine it by calculating the moles of Mg(OH)2 produced. The one that will produce the SMALLER amount is limiting.
MgCl2 to Mg(OH)2:
0.0481 mol MgCl2 * (1 mol Mg(OH)2 / 1 mol MgCl2) = 0.0481 mol Mg(OH)2
NaOH to Mg(OH)2:
0.02665 mol NaOH * (1 mol Mg(OH)2 / 2 mol NaOH) = 0.01333 mol Mg(OH)2
Thus, NaOH is limiting. You will base you calculation of the mass of Mg(OH)2 on the moles produced by NaOH. To get the mass, just multiply the molar mass of Mg(OH)2 by the moles we got (which is 0.01333 mol Mg(OH)2).
Hope this helps~ `u`
148 mL of 0.325 M MgCl2 is mixed with 68.7 mL of 0.388 M NaOH and allowed to react completely. how many grams of Mg(OH)2 will form?
1 answer