How many grams of magnesium hydroxide will precipitate if 25.0 mL of 0.235 M magnesium nitrate are combined with 30 mL of 0.260 M potassium hydroxide?
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To find the number of grams of magnesium hydroxide that will precipitate, we need to determine the limiting reactant and calculate the amount of magnesium hydroxide formed based on that.
Step 1: Write the balanced chemical equation for the reaction between magnesium nitrate and potassium hydroxide.
Mg(NO3)2 + 2KOH → Mg(OH)2 + 2KNO3
Step 2: Calculate the moles of magnesium nitrate.
Moles of Mg(NO3)2 = volume (L) × concentration (M)
Moles of Mg(NO3)2 = 25.0 mL × 0.235 M
Moles of Mg(NO3)2 = 0.0250 L × 0.235 mol/L
Moles of Mg(NO3)2 = 0.00588 mol
Step 3: Calculate the moles of potassium hydroxide.
Moles of KOH = volume (L) × concentration (M)
Moles of KOH = 30 mL × 0.260 M
Moles of KOH = 0.0300 L × 0.260 mol/L
Moles of KOH = 0.00780 mol
Step 4: Determine the limiting reactant.
The balanced chemical equation tells us that the molar ratio of Mg(NO3)2 to Mg(OH)2 is 1 to 1. The molar ratio of KOH to Mg(OH)2 is 2 to 1. Therefore, we need twice as many moles of KOH as Mg(NO3)2 to react completely.
Since the molar ratio of KOH to Mg(NO3)2 is 2 to 1 and we have fewer moles of KOH than Mg(NO3)2, KOH is the limiting reactant.
Step 5: Calculate the moles of magnesium hydroxide formed.
From the balanced equation, we know that the molar ratio of KOH to Mg(OH)2 is 2 to 1. Therefore, the moles of Mg(OH)2 formed will be half the moles of KOH used.
Moles of Mg(OH)2 = 0.00780 mol KOH / 2
Moles of Mg(OH)2 = 0.00390 mol
Step 6: Calculate the grams of magnesium hydroxide formed.
Grams of Mg(OH)2 = moles × molar mass
Grams of Mg(OH)2 = 0.00390 mol × 58.32 g/mol
Grams of Mg(OH)2 ≈ 0.227 g
Therefore, approximately 0.227 grams of magnesium hydroxide will precipitate.
To determine the number of grams of magnesium hydroxide that will precipitate, we need to calculate the limiting reactant in the reaction between magnesium nitrate (Mg(NO3)2) and potassium hydroxide (KOH). The balanced chemical equation for the reaction is:
Mg(NO3)2 + 2KOH -> Mg(OH)2 + 2KNO3
To find the limiting reactant, we first need to calculate the number of moles of each reactant. We can use the equation:
moles = concentration (M) * volume (L)
For magnesium nitrate (Mg(NO3)2):
moles of Mg(NO3)2 = 0.235 M * 0.0250 L = 0.00587 mol
For potassium hydroxide (KOH):
moles of KOH = 0.260 M * 0.0300 L = 0.00780 mol
According to the balanced equation, the stoichiometric ratio between Mg(NO3)2 and Mg(OH)2 is 1:1. Therefore, the number of moles of magnesium hydroxide (Mg(OH)2) formed will be equal to the number of moles of magnesium nitrate (Mg(NO3)2).
moles of Mg(OH)2 = 0.00587 mol
Now, we can calculate the mass of magnesium hydroxide (Mg(OH)2) using its molar mass:
Molar mass of Mg(OH)2 = 24.31 g/mol + 2 * 1.01 g/mol + 2 * 16.00 g/mol = 58.31 g/mol
mass of Mg(OH)2 = moles of Mg(OH)2 * molar mass of Mg(OH)2
= 0.00587 mol * 58.31 g/mol
Calculating the result:
mass of Mg(OH)2 = 0.342 g
Therefore, approximately 0.342 grams of magnesium hydroxide will precipitate in the reaction.