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?
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`
To find the number of grams of Mg(OH)2 that will form, we must first determine the balanced chemical equation for the reaction between MgCl2 and NaOH.
The chemical equation for the reaction can be written as follows:
MgCl2 + 2NaOH -> Mg(OH)2 + 2NaCl
From the balanced equation, we can see that one mole of MgCl2 reacts with two moles of NaOH to produce one mole of Mg(OH)2.
To calculate the number of moles of MgCl2 and NaOH, we need to use the equation:
moles = concentration (in M) * volume (in L)
For MgCl2:
moles of MgCl2 = 0.325 M * 0.148 L = 0.0479 moles
For NaOH:
moles of NaOH = 0.388 M * 0.0687 L = 0.0267 moles
Since the reaction is 1:1 between MgCl2 and Mg(OH)2, the number of moles of Mg(OH)2 formed will be equal to the number of moles of MgCl2.
Therefore, the number of moles of Mg(OH)2 that will form is 0.0479 moles.
Now, we need to calculate the molar mass of Mg(OH)2. The molar mass is calculated by adding up the atomic masses of each element in the compound.
Molar mass of Mg(OH)2 = (1 x atomic mass of Mg) + (2 x atomic mass of O) + (2 x atomic mass of H)
= (1 x 24.31) + (2 x 16.00) + (2 x 1.01)
= 58.33 g/mol
Finally, to find the mass of Mg(OH)2, we can use the equation:
mass = moles * molar mass
mass of Mg(OH)2 = 0.0479 moles * 58.33 g/mol = 2.79 grams
Therefore, approximately 2.79 grams of Mg(OH)2 will form.