MgCl2(aq) + 2 NaOH(aq) --> Mg(OH)2(s) + 2 NaCl(aq)
How many grams of Mg(OH)2 would form if 50.00 mL of 0.9750M NaOH were combined with 50.00 mL of 0.8820 M MgCl2?
To determine the grams of Mg(OH)2 that would form, we need to follow these steps:
Step 1: Calculate the moles of NaOH and MgCl2
To find the moles of NaOH:
moles of NaOH = volume (in liters) × concentration (in M)
Since the volume is given in mL, we need to convert it to liters by dividing by 1000:
volume of NaOH = 50.00 mL ÷ 1000 = 0.0500 L
moles of NaOH = 0.0500 L × 0.9750 M = 0.04875 moles
To find the moles of MgCl2:
moles of MgCl2 = volume (in liters) × concentration (in M)
volume of MgCl2 = 50.00 mL ÷ 1000 = 0.0500 L
moles of MgCl2 = 0.0500 L × 0.8820 M = 0.0441 moles
Step 2: Determine the limiting reactant
The limiting reactant is the one that is completely consumed and determines the amount of product formed. To find the limiting reactant, we compare the moles of NaOH and MgCl2.
From the balanced equation, we can see that the stoichiometric ratio between NaOH and Mg(OH)2 is 2:1. Therefore, we need to compare the moles of NaOH and MgCl2 using these ratios.
moles of NaOH:moles of MgCl2 = 0.04875 moles : 0.0441 moles
Simplifying, we get:
moles of NaOH:moles of MgCl2 ≈ 1.105 : 1
Since the ratio is approximately 1:1, the limiting reactant is MgCl2.
Step 3: Calculate the moles of Mg(OH)2
Since MgCl2 is the limiting reactant, the moles of Mg(OH)2 formed will be equal to the moles of MgCl2.
moles of Mg(OH)2 = 0.0441 moles
Step 4: Calculate the mass of Mg(OH)2
To find the mass of Mg(OH)2, we multiply the moles by the molar mass of Mg(OH)2. The molar mass of Mg(OH)2 is the sum of the atomic masses of one Mg atom, two O atoms, and two H atoms.
molar mass of Mg(OH)2 = (24.31 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol)
molar mass of Mg(OH)2 = 58.33 g/mol
mass of Mg(OH)2 = moles of Mg(OH)2 × molar mass of Mg(OH)2
mass of Mg(OH)2 = 0.0441 moles × 58.33 g/mol
Therefore, the grams of Mg(OH)2 that would form is approximately 2.57 grams.
Calculate moles MgCl2 to start = M x L = ?
Calculate moles NaOH to start = M x L = ?
This is a limiting reagent problem; I know that because amounts for BOTH reactants are given. I solve these by solving two simple stoichiometry problems.
First moles MgCl2, ignoring NaOH, second with moles NaOH, ignoring MgCl2.
Convert moles MgCl2 to moles of the product using the coefficients in the balanced equation. Do the same for NaOH to moles of the product. Both answers can't be right; the correct one is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Then convert moles of the product (using the smaller value) to grams. g = moles x molar mass.
Here is a worked example of a stoichiometry problem. This will solve all of your stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html