60.0 mL of 3.40 M sodium hydroxide is combined with 39.0 mL of 1.50 M magnesium chloride. What mass in grams of solid forms? (Hint: the total or overall reaction is most useful in doing calculations.)
I got 485.1
2NaOH + MgCl2 ==> Mg(OH)2(solid) + 2HCl
This is a limiting reagent problem. I know that because amounts for BOTH reactants are given.
mols NaOH = M x L = ?
mols MgCl2 = M x L = ?
Convert mols NaOH to mols product using the coefficients in the balanced equation.
Do the same for mols MgCl2.
It is likely that the two numbers will not be the same which means one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
grams Mg(OH)2 = mols x molar mass.
Okay so I followed what you said and got 78. Does that make sense?
No. I don't get that.
I will work in millimoles and convert mols.
mL x M NaOH = 3.4 x 60 = 204 millimoles = 0.204 mols.
mL x M MgCl2 = 39.0 x 1.50 M = 58.5 mmols = 0.0585 mols.
Convert 0.204 mols NaOH to mols Mg(OH)2. That's 0.204 x (1 mol Mg(OH)2/2 mols NaOH) = 0.204 x 1/2 = 0.102 mols Mg(OH)2.
Convert 0.0585 mols MgCl2 to mols Mg(OH)2.
0.0585 x (1 mol Mg(OH)2/1 mol MgCl2) = 0.0585 x 1/1 = 0.0585 mol Mg(OH)2
The small number wins; 0.0585 mols Mg(OH)2 formed and MgCl2 is the limiting reagent.
g Mg(OH)2 = mols x molar mass = 0.0585 x 58.32 = 3.41 g.
Thanks! That makes sense, and the answer was right!
To determine the mass of the solid that forms when sodium hydroxide (NaOH) and magnesium chloride (MgCl2) react, you need to first write the balanced chemical equation for the reaction.
The balanced equation for the reaction between sodium hydroxide and magnesium chloride is as follows:
2NaOH(aq) + MgCl2(aq) → 2NaCl(aq) + Mg(OH)2(s)
From the balanced equation, it can be seen that the mole ratio between NaOH and Mg(OH)2 is 2:1. This means that for every 2 moles of NaOH, 1 mole of Mg(OH)2 is formed.
Now, let's calculate the number of moles of NaOH and MgCl2 in the given volumes of their respective solutions.
Moles of NaOH:
Volume of NaOH solution = 60.0 mL = 0.0600 L (since 1 mL = 0.001 L)
Molarity of NaOH solution = 3.40 M
Moles of NaOH = Volume (in L) × Molarity
= 0.0600 L × 3.40 mol/L
= 0.204 mol
Moles of MgCl2:
Volume of MgCl2 solution = 39.0 mL = 0.0390 L (since 1 mL = 0.001 L)
Molarity of MgCl2 solution = 1.50 M
Moles of MgCl2 = Volume (in L) × Molarity
= 0.0390 L × 1.50 mol/L
= 0.0585 mol
According to the balanced equation, the ratio of moles between NaOH and Mg(OH)2 is 2:1. Since we have 0.204 moles of NaOH, the number of moles of Mg(OH)2 formed will be half of that:
Moles of Mg(OH)2 = 0.204 mol ÷ 2
= 0.102 mol
Finally, to determine the mass of solid Mg(OH)2 formed, we need to use its molar mass.
The molar mass of Mg(OH)2 can be calculated by adding the atomic masses of its constituent elements:
Mg: 24.31 g/mol
O: 16.00 g/mol (2 atoms)
H: 1.01 g/mol (2 atoms)
Molar mass of Mg(OH)2 = 24.31 g/mol + (16.00 g/mol × 2) + (1.01 g/mol × 2)
= 24.31 g/mol + 32.00 g/mol + 2.02 g/mol
= 58.33 g/mol
Mass of Mg(OH)2 formed = Moles of Mg(OH)2 × Molar mass of Mg(OH)2
= 0.102 mol × 58.33 g/mol
= 5.956 g
Therefore, the mass of the solid Mg(OH)2 formed when 60.0 mL of 3.40 M NaOH is combined with 39.0 mL of 1.50 M MgCl2 is approximately 5.956 grams.