When copper(II) nitrate reacts with sodium hydroxide, copper (II) hydroxide is produced. How many grams of Cu(OH)2 can be prepared from 12.7g of Cu(NO3)2 and excess NaOH?

This four step process will work all of your stoichiometry problems. Copy and save this.

1. Write and balance the equation.
Cu(NO3)2 + 2NaOH ==> Cu(OHN)2 + 2NaNO3

2. Convert grams of what you have to mols.
mols = grams/molar mass = ?

3. Using the coefficients in the balanced equation, convert mols of what you have [in this case mols Cu(NO3)2] to mols of what you want [in this case Cu(OH)2].

4. Now convert mols Cu(OH)2 to grams.
grams = mols x molar mass

Post your work if you get stuck.

What are the answers?

Why did the chemist go broke?

Because he couldn't find any solutions!

Now, let's calculate the amount of copper(II) hydroxide that can be prepared from 12.7g of copper(II) nitrate.

The molar mass of Cu(NO3)2 is:
Cu: 63.55 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (x3)

Total molar mass of Cu(NO3)2 = (63.55) + (14.01) + (16.00 x3) = 187.55 g/mol

To find the molar amount of Cu(NO3)2, divide the given mass by the molar mass:
Moles of Cu(NO3)2 = 12.7 g / 187.55 g/mol

Now, let's look at the balanced chemical equation:
Cu(NO3)2 + 2NaOH → Cu(OH)2 + 2NaNO3

From the equation, we can see that 1 mole of Cu(NO3)2 produces 1 mole of Cu(OH)2. Therefore, the molar ratio between Cu(NO3)2 and Cu(OH)2 is 1:1.

This means that the moles of Cu(OH)2 is the same as the moles of Cu(NO3)2.

So, the moles of Cu(OH)2 = 12.7 g / 187.55 g/mol

Finally, to find the mass of Cu(OH)2, multiply the moles by the molar mass of Cu(OH)2:
Mass of Cu(OH)2 = moles of Cu(OH)2 x molar mass of Cu(OH)2

I hope this helps you calculate how many grams of Cu(OH)2 can be prepared! Just remember, always wear safety goggles while handling chemicals. Safety first, laughter second!

To determine the amount of Cu(OH)2 that can be prepared, we need to first balance the equation:

Cu(NO3)2 + 2NaOH -> Cu(OH)2 + 2NaNO3

From the balanced equation, we can see that 1 mole of Cu(NO3)2 reacts with 2 moles of NaOH to produce 1 mole of Cu(OH)2.

Now let's calculate the moles of Cu(NO3)2:

molar mass of Cu(NO3)2 = atomic mass of Cu + 2 * (atomic mass of N + 3 * atomic mass of O)
= 63.55 g/mol + 2 * (14.01 g/mol + 3 * 16.00 g/mol)
= 63.55 g/mol + 2 * (14.01 g/mol + 48.00 g/mol)
= 63.55 g/mol + 2 * (62.01 g/mol)
= 63.55 g/mol + 124.02 g/mol
= 187.57 g/mol

moles of Cu(NO3)2 = mass of Cu(NO3)2 / molar mass of Cu(NO3)2
= 12.7 g / 187.57 g/mol
≈ 0.0677 mol (rounded to four decimal places)

Since the reaction is 1:1 between Cu(NO3)2 and Cu(OH)2, the moles of Cu(OH)2 will also be 0.0677 mol.

Next, we need to calculate the mass of Cu(OH)2:

molar mass of Cu(OH)2 = atomic mass of Cu + 2 * atomic mass of O + 2 * atomic mass of H
= 63.55 g/mol + 2 * 16.00 g/mol + 2 * 1.01 g/mol
= 63.55 g/mol + 32.00 g/mol + 2.02 g/mol
= 97.57 g/mol

mass of Cu(OH)2 = moles of Cu(OH)2 * molar mass of Cu(OH)2
= 0.0677 mol * 97.57 g/mol
≈ 6.57 g

Therefore, approximately 6.57 grams of Cu(OH)2 can be prepared from 12.7 grams of Cu(NO3)2 and excess NaOH.

To determine the number of grams of Cu(OH)2 that can be produced, we need to calculate the theoretical yield of the reaction. The theoretical yield represents the maximum amount of product that can be obtained under ideal conditions.

Let's start by writing the balanced chemical equation for the reaction:

Cu(NO3)2 + 2NaOH -> Cu(OH)2 + 2NaNO3

From the equation, we can see that one mole of Cu(NO3)2 reacts with two moles of NaOH to produce one mole of Cu(OH)2. Therefore, we need to convert grams of Cu(NO3)2 to moles, and then use the stoichiometry to find the moles of Cu(OH)2 produced.

1. Convert grams of Cu(NO3)2 to moles:
To do this, we need to know the molar mass of Cu(NO3)2. The molar mass of copper (Cu) is 63.55 g/mol, nitrogen (N) is 14.01 g/mol, and oxygen (O) is 16.00 g/mol.

Cu(NO3)2 molar mass = 1 * Cu molar mass + 2 * (N molar mass + 3 * O molar mass)
= 1 * 63.55 g/mol + 2 * (14.01 g/mol + 3 * 16.00 g/mol)
= 63.55 g/mol + 2 * (14.01 g/mol + 48.00 g/mol)
= 63.55 g/mol + 2 * 62.01 g/mol
= 63.55 g/mol + 124.02 g/mol
= 187.57 g/mol

Now, we can calculate the number of moles of Cu(NO3)2:
mols of Cu(NO3)2 = mass / molar mass
= 12.7 g / 187.57 g/mol
≈ 0.068 moles (rounded to three decimal places)

2. Use stoichiometry to find moles of Cu(OH)2:
From the balanced equation, we know that one mole of Cu(NO3)2 produces one mole of Cu(OH)2.
Therefore, moles of Cu(OH)2 = moles of Cu(NO3)2 = 0.068 moles (from step 1).

3. Convert moles of Cu(OH)2 to grams:
Again, we need the molar mass of Cu(OH)2. The molar mass of Cu is 63.55 g/mol, O is 16.00 g/mol, and H is 1.01 g/mol.

Cu(OH)2 molar mass = Cu molar mass + 2 * (O molar mass + H molar mass)
= 63.55 g/mol + 2 * (16.00 g/mol + 1.01 g/mol)
= 63.55 g/mol + 2 * 17.01 g/mol
= 63.55 g/mol + 34.02 g/mol
= 97.57 g/mol

Now, we can calculate the mass of Cu(OH)2:
mass = moles of Cu(OH)2 * molar mass
= 0.068 moles * 97.57 g/mol
≈ 6.64 grams (rounded to two decimal places)

Therefore, approximately 6.64 grams of Cu(OH)2 can be prepared from 12.7 grams of Cu(NO3)2 and excess NaOH.