1) Write a balanced chemical equation for the reaction between Cu(NO3)2 * 3 H2O and NaOH. Underline the formula for the precipitate produced by this reaction. (The water of hydration in Cu(NO3)2 * 3 H2O appears as liquid water on the right side of the equation)
2) Calculate the molar masses of copper(II) nitrate trihydrate and of the precipitate formed by this reaction.
3)Calculate the number of moles of NaOh in 5.00 mL of a 1.00 M solution of NaOH.
*Please help with all questions!*
1. Cu(NO3)2.3H2O(aq) + NaOH(aq) --> Cu(OH)2(s) + H2O + NaNO3(aq)
We can't underline but (s) means a solid ppt. You will need to balance it.
2. This is a stoichiometry problem. You have worked many like it.
3. moles = M x L.
1) To write a balanced chemical equation for the reaction between Cu(NO3)2 * 3 H2O (copper(II) nitrate trihydrate) and NaOH, we first need to know the formulas of the products. In this reaction, copper(II) hydroxide will be formed as the precipitate (solid). The balanced chemical equation is as follows:
Cu(NO3)2 * 3 H2O + 2 NaOH -> Cu(OH)2 + 2 NaNO3 + 3 H2O
The formula for the precipitate, copper(II) hydroxide (Cu(OH)2), is underlined.
2) To calculate the molar mass of Cu(NO3)2 * 3 H2O, we need to find the molar mass of each element and sum them up.
Copper (Cu) has a molar mass of 63.55 g/mol.
Nitrogen (N) has a molar mass of 14.01 g/mol.
Oxygen (O) has a molar mass of 16.00 g/mol.
Hydrogen (H) has a molar mass of 1.01 g/mol.
For copper nitrate (Cu(NO3)2), we have:
Cu: 1 * 63.55 g/mol = 63.55 g/mol
N: 2 * 14.01 g/mol = 28.02 g/mol
O: 6 * 16.00 g/mol = 96.00 g/mol
For water (H2O), we have:
H: 6 * 1.01 g/mol = 6.06 g/mol
O: 3 * 16.00 g/mol = 48.00 g/mol
By adding up the molar masses of each element, we get:
Cu(NO3)2 * 3 H2O: 63.55 + 28.02 + 96.00 + 6.06 + 48.00 = 241.63 g/mol
The molar mass of the precipitate, copper(II) hydroxide (Cu(OH)2), can be calculated similarly:
Cu: 1 * 63.55 g/mol = 63.55 g/mol
O: 2 * 16.00 g/mol = 32.00 g/mol
H: 2 * 1.01 g/mol = 2.02 g/mol
Cu(OH)2: 63.55 + 32.00 + 2.02 = 97.57 g/mol
3) To calculate the number of moles of NaOH, we need to use the formula:
moles = volume (in liters) * concentration (in moles/liter)
Given:
Volume = 5.00 mL = 5.00 * 10^(-3) L
Concentration = 1.00 M
Using the formula, we have:
moles = 5.00 * 10^(-3) L * 1.00 mol/L = 5.00 * 10^(-3) mol
Thus, there are 5.00 * 10^(-3) moles of NaOH in 5.00 mL of a 1.00 M solution.
1) To write a balanced chemical equation for the reaction between Cu(NO3)2 * 3 H2O and NaOH, we need to determine the formulas of the reactants and products.
The formula for copper(II) nitrate trihydrate is Cu(NO3)2 * 3 H2O. This means that for every one copper(II) nitrate trihydrate compound, there are three water molecules present.
The formula for sodium hydroxide is NaOH.
To balance the equation, we start by separating the reactants and products:
Cu(NO3)2 * 3 H2O + 2 NaOH →
Now, let's balance the atoms starting with the copper (Cu) atoms. We have one Cu on the left-hand side, so we'll need one Cu on the right-hand side:
Cu(NO3)2 * 3 H2O + 2 NaOH → Cu +
Next, let's balance the nitrogen (N) atoms. We have two N atoms in the nitrate (NO3)2 on the left, so we'll need two N atoms on the right:
Cu(NO3)2 * 3 H2O + 2 NaOH → Cu + 2 NaNO3 +
Now, let's balance the hydrogen (H) atoms. On the left, we have six H atoms from the three water molecules (H2O), so we'll need six H atoms on the right:
Cu(NO3)2 * 3 H2O + 2 NaOH → Cu + 2 NaNO3 + 6 H2O
Finally, let's balance the oxygen (O) atoms. On the left, we have six O atoms from the nitrate ions (NO3)2, and six O atoms from the water molecules (H2O). On the right, we have two O atoms from the sodium hydroxide (NaOH) and nine O atoms from the nitrate ions (NO3)3:
Cu(NO3)2 * 3 H2O + 2 NaOH → Cu + 2 NaNO3 + 6 H2O
The balanced chemical equation for the reaction between Cu(NO3)2 * 3 H2O and NaOH is:
Cu(NO3)2 * 3 H2O + 2 NaOH → Cu + 2 NaNO3 + 6 H2O
The precipitate produced in this reaction is Cu(OH)2, which is copper(II) hydroxide.
2) To calculate the molar mass of copper(II) nitrate trihydrate (Cu(NO3)2 * 3 H2O), we need to find the molar masses of each of its individual elements and calculate the total.
The molar masses are as follows:
- Atomic mass of Cu = 63.55 g/mol
- Atomic mass of N = 14.01 g/mol
- Atomic mass of O = 16.00 g/mol
- Atomic mass of H = 1.01 g/mol
The formula has one Cu atom, two N atoms, and nine O atoms, along with three water molecules, which contain three H2O units.
Therefore, the molar mass of Cu(NO3)2 * 3 H2O can be calculated as follows:
(1 * 63.55 g/mol) + (2 * 14.01 g/mol) + (9 * 16.00 g/mol) + (3 * (2 * 1.01 g/mol))
= 63.55 g/mol + 28.02 g/mol + 144.00 g/mol + 6.