An aqueous solution of potassium hydroxide is mixed with an (aq) of magnesium chloride and you get a clumpy white precipitate that forms. Write a complete and balanced equation to calculate the stoichiometry question: how many grams of the precipitate are formed if 150 grams of (aq) potassium hydroxide are used as a reactant? Consider magnesium chloride to be an excess reactant that we need not quantify. Be sure to include the stat of the reactant and products in your balanced equation.
MgCl2(aq) + 2KOH(aq) ==> 2KCl(aq) + Mg(OH)2(s)
Here is an example stoichiometry problem worked. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Sure! But before I dive into the equation, let me throw in a little chemistry humor. Why don't scientists trust atoms? Because they make up everything!
Now, let's get back to your question. The balanced equation for the reaction is:
2KOH(aq) + MgCl2(aq) → 2KCl(aq) + Mg(OH)2(s)
Now, to calculate the grams of precipitate formed, we need to determine the stoichiometry of the reaction. From the balanced equation, we see that 2 moles of KOH react with 1 mole of MgCl2 to form 1 mole of Mg(OH)2.
To solve this, we need to convert grams of KOH to moles and then to grams of Mg(OH)2. Let's do the math!
The molar mass of KOH is 56.11 g/mol. Therefore, 150 grams of KOH is equal to:
150 g / 56.11 g/mol = 2.675 moles of KOH
Since the stoichiometric ratio between KOH and Mg(OH)2 is 2:1, we have:
2.675 moles of KOH → 1.3375 moles of Mg(OH)2
Now, we need to convert moles of Mg(OH)2 to grams. The molar mass of Mg(OH)2 is 58.32 g/mol. So, 1.3375 moles of Mg(OH)2 is equal to:
1.3375 moles × 58.32 g/mol = 77.996 g
Therefore, approximately 78 grams of the clumpy white precipitate (Mg(OH)2) will be formed when 150 grams of KOH is used as a reactant.
Remember, though, this is just an approximate value, as it assumes 100% yield and complete conversion of reactants. In reality, factors like purity, side reactions, and efficiency may affect the actual yield.
The balanced equation for the reaction between potassium hydroxide (KOH) and magnesium chloride (MgCl2) is:
2 KOH(aq) + MgCl2(aq) → Mg(OH)2(s) + 2 KCl(aq)
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of MgCl2 to form 1 mole of magnesium hydroxide (Mg(OH)2).
To calculate the grams of the precipitate formed, we need to find the molar mass of Mg(OH)2. The molar mass of Mg = 24.31 g/mol, O = 16.00 g/mol, and H = 1.01 g/mol.
Molar mass of Mg(OH)2 = 24.31 g/mol + (16.00 g/mol * 2) + (1.01 g/mol * 2) = 58.33 g/mol
Now, we can calculate the moles of the precipitate formed using the given mass of KOH.
Moles of KOH = mass / molar mass = 150 g / 56.11 g/mol = 2.674 mol
According to the balanced equation, the stoichiometry between KOH and Mg(OH)2 is 2:1. Therefore, the moles of Mg(OH)2 formed will be half of the moles of KOH used.
Moles of Mg(OH)2 = 2.674 mol / 2 = 1.337 mol
Finally, we can calculate the mass of the precipitate formed using the moles and molar mass.
Mass of precipitate = moles of Mg(OH)2 * molar mass of Mg(OH)2
= 1.337 mol * 58.33 g/mol
= 77.98 grams
Therefore, approximately 77.98 grams of the precipitate will be formed if 150 grams of potassium hydroxide are used as a reactant.
To write the balanced equation for this reaction, we need to first identify the chemical formulas of the substances involved.
The chemical formula for potassium hydroxide is KOH, and the chemical formula for magnesium chloride is MgCl2.
The balanced equation for the reaction between potassium hydroxide and magnesium chloride can be represented as:
2 KOH(aq) + MgCl2(aq) → 2 KCl(aq) + Mg(OH)2(s)
In this equation, the aqueous solutions of potassium hydroxide and magnesium chloride react to form an aqueous solution of potassium chloride and a solid precipitate of magnesium hydroxide.
Now, let's calculate the number of grams of the precipitate formed if 150 grams of aqueous potassium hydroxide is used.
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of Mg(OH)2. So, we can use the molar mass of KOH to determine the number of moles of KOH used.
The molar mass of KOH is 39.10 g/mol + 16.00 g/mol + 1.01 g/mol = 56.11 g/mol.
Using the given mass of 150 grams of KOH, we can calculate the number of moles as follows:
Number of moles of KOH = Mass of KOH / Molar mass of KOH
= 150 g / 56.11 g/mol
≈ 2.67 mol
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of Mg(OH)2. Therefore, 2.67 moles of KOH will react to form an equal number of moles of Mg(OH)2.
Thus, the number of moles of Mg(OH)2 formed = 2.67 moles.
Now, to calculate the mass of the precipitate, we need to multiply the number of moles of Mg(OH)2 by its molar mass.
The molar mass of Mg(OH)2 is 24.31 g/mol + (16.00 g/mol + 1.01 g/mol) × 2 = 58.33 g/mol.
Mass of the precipitate = Number of moles of Mg(OH)2 × Molar mass of Mg(OH)2
= 2.67 mol × 58.33 g/mol
≈ 155.5 grams
Therefore, approximately 155.5 grams of the precipitate are formed when 150 grams of aqueous potassium hydroxide is reacted with an excess of magnesium chloride.