copper sulfate reacts with sodium hydroxide producing copper hydroxide and sodium sulfate.. If sodium hydroxide is the limiting agent how many grams of copper hydroxide should precipitate out?{There are 634.44 g of copper sulfate and 240 grams of sodium hydroxide.

To determine how many grams of copper hydroxide will precipitate out when sodium hydroxide is the limiting agent, we need to calculate the amount of copper hydroxide that can be formed from the reaction.

Let's start by writing the balanced chemical equation for the reaction:
CuSO4 + 2 NaOH → Cu(OH)2 + Na2SO4

According to the equation, 1 mole of copper sulfate (CuSO4) reacts with 2 moles of sodium hydroxide (NaOH) to yield 1 mole of copper hydroxide (Cu(OH)2) and 1 mole of sodium sulfate (Na2SO4).

Convert the given mass of copper sulfate to moles:
molar mass of CuSO4 = 63.55 g/mol (copper) + 32.06 g/mol (sulfur) + (4 * 16.00 g/mol) (oxygen)
molar mass of CuSO4 = 159.60 g/mol

moles of CuSO4 = mass / molar mass
moles of CuSO4 = 634.44 g / 159.60 g/mol
moles of CuSO4 = 3.98 mol

From the balanced equation, 1 mole of CuSO4 reacts to form 1 mole of Cu(OH)2.

Therefore, moles of Cu(OH)2 = 3.98 mol

Now, let's calculate the moles of sodium hydroxide (NaOH). We will compare this value to the available moles of Cu(OH)2 to determine the limiting agent.

moles of NaOH = mass / molar mass
moles of NaOH = 240 g / 40.00 g/mol
moles of NaOH = 6 mol

From the balanced equation, 2 moles of NaOH react to form 1 mole of Cu(OH)2.

Using the mole ratio, we find that the expected moles of Cu(OH)2 from the reaction with NaOH is (6 mol NaOH / 2 mol NaOH) = 3 mol Cu(OH)2.

Since we have 3.98 mol Cu(OH)2 available (from CuSO4), which is more than the expected 3 mol (from NaOH), the sodium hydroxide is the limiting agent.

Finally, to find the mass of Cu(OH)2 that will precipitate out when NaOH is the limiting agent, we need to convert moles to grams using the molar mass of Cu(OH)2.

molar mass of Cu(OH)2 = (63.55 g/mol (copper) + 2 * (16.00 g/mol (oxygen) + 1.01 g/mol (hydrogen))
molar mass of Cu(OH)2 = 97.56 g/mol

mass of Cu(OH)2 = moles of Cu(OH)2 * molar mass
mass of Cu(OH)2 = 3 mol * 97.56 g/mol
mass of Cu(OH)2 = 292.68 g

Therefore, approximately 292.68 grams of copper hydroxide should precipitate out when sodium hydroxide is the limiting agent.

To determine how many grams of copper hydroxide will precipitate out, we need to follow these steps:

Step 1: Write and balance the chemical equation:
CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

Step 2: Calculate the molar mass of copper sulfate (CuSO4):
Cu: 63.55 g/mol
S: 32.06 g/mol
O: 16.00 g/mol

Molar mass of CuSO4 = (63.55 g/mol) + (32.06 g/mol) + (4 * 16.00 g/mol) = 159.61 g/mol

Step 3: Calculate the number of moles of copper sulfate (CuSO4):
Number of moles = mass / molar mass
Number of moles of CuSO4 = 634.44 g / 159.61 g/mol = 3.976 mol

Step 4: Determine the stoichiometric ratio of CuSO4 to Cu(OH)2:
The stoichiometric ratio can be found from the balanced equation. For every 1 mole of CuSO4, 1 mole of Cu(OH)2 is produced.

Step 5: Calculate the number of moles of Cu(OH)2 that will be produced:
Since the stoichiometric ratio is 1:1, the number of moles of Cu(OH)2 will be the same as the number of moles of CuSO4.

Moles of Cu(OH)2 = 3.976 mol

Step 6: Calculate the molar mass of copper hydroxide (Cu(OH)2):
Cu: 63.55 g/mol
O: 16.00 g/mol
H: 1.01 g/mol

Molar mass of Cu(OH)2 = (63.55 g/mol) + (2 * 16.00 g/mol) + (2 * 1.01 g/mol) = 97.55 g/mol

Step 7: Calculate the mass of copper hydroxide that will be precipitated out:
Mass = moles x molar mass

Mass of Cu(OH)2 = 3.976 mol x 97.55 g/mol ≈ 387.88 g

Therefore, approximately 387.88 grams of copper hydroxide should precipitate out.