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.
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
4 answers
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.
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!
1 answer