What is the limiting reactant in this experiment? Assume you used 1.0 g of the copper(II) sulfate pentahydrate and 7. 5 mL of 6.0 M NH3. Show your calculations.

Its been a while need some help, thanks!

To determine the limiting reactant in the experiment, we need to compare the amount of each reactant (copper(II) sulfate pentahydrate and NH3) to their respective stoichiometric ratios in the balanced chemical equation.

The balanced chemical equation for the reaction between copper(II) sulfate pentahydrate (CuSO4 · 5H2O) and ammonia (NH3) is:

CuSO4 · 5H2O + 4NH3 → Cu(NH3)4SO4 + 5H2O

To make the comparison, we'll convert the mass of copper(II) sulfate pentahydrate (CuSO4 · 5H2O) and the volume of ammonia (NH3) solution to moles.

First, let's calculate the moles of copper(II) sulfate pentahydrate (CuSO4 · 5H2O):

Molar mass of CuSO4 · 5H2O:
Cu: 63.55 g/mol
S: 32.07 g/mol
O: 16.00 g/mol
H: 1.01 g/mol

Total molar mass = (63.55 + 32.07 + (16.00 x 4) + (1.01 x 2 x 5)) g/mol = 250.68 g/mol

Moles of CuSO4 · 5H2O = (1.0 g) / (250.68 g/mol)

Next, let's calculate the moles of ammonia (NH3):

Molarity of NH3 solution = 6.0 M
Volume of NH3 solution = 7.5 mL = 7.5 x 10^(-3) L

Moles of NH3 = (6.0 mol/L) x (7.5 x 10^(-3) L)

Now, we'll use the stoichiometric ratios from the balanced chemical equation to determine the limiting reactant.

From the balanced equation, the ratio of moles of CuSO4 · 5H2O to moles of NH3 is 1:4. Therefore, if the ratio of moles of CuSO4 · 5H2O to moles of NH3 is less than 1:4, then CuSO4 · 5H2O is the limiting reactant. Otherwise, if the ratio is greater than 1:4, then NH3 is the limiting reactant.

Let's calculate the ratio:

Ratio = (moles of CuSO4 · 5H2O) / (moles of NH3)

If the ratio is less than 1:4, CuSO4 · 5H2O is the limiting reactant.
If the ratio is greater than 1:4, NH3 is the limiting reactant.

Please provide the value for the moles of CuSO4 · 5H2O and moles of NH3 calculated above so that we can determine the limiting reactant.

To determine the limiting reactant in this experiment, we need to compare the amount of product that can be produced using each reactant. The reactant that produces the least amount of product is the limiting reactant.

First, we need to calculate the number of moles for each reactant:

1. For the copper(II) sulfate pentahydrate:
Given mass: 1.0 g
Molar mass of CuSO4·5H2O: 159.609 g/mol (you can find this on the periodic table)
Number of moles = mass / molar mass
Number of moles = 1.0 g / 159.609 g/mol

2. For the NH3:
Given volume: 7.5 mL
Molarity of NH3: 6.0 M (moles per liter)
Number of moles = volume (in liters) x molarity
Converting mL to L: 7.5 mL / 1000 mL/L
Number of moles = (7.5 / 1000) L x 6.0 mol/L

Now that we have the number of moles for each reactant, we can use the balanced chemical equation to determine the stoichiometry:

CuSO4·5H2O + 2NH3 -> Cu(NH3)2SO4 + 5H2O

From the balanced equation, we can see that 1 mole of CuSO4·5H2O reacts with 2 moles of NH3.

Next, we compare the number of moles of each reactant to find the limiting reactant:

Divide the moles of each reactant by their respective stoichiometric coefficients:
Moles of CuSO4·5H2O / 1 = Moles of NH3 / 2

Now you can substitute the calculated values:

(1.0 g / 159.609 g/mol) / 1 = [(7.5 / 1000) L x 6.0 mol/L] / 2

Simplify the equation above to find the limiting reactant. The reactant with the smaller value is the limiting reactant.

And what is the experiment? What is the reaction? Is this it?

CuSO4.5H2O + 4NH3 --> [Cu(NH3)4]SO4.5H2O
mols CuSO4.5H2O = grams/molar mass = ?
mols NH3 = M x L = ?

Now convert mols each to mols product using the coefficients in the balanced equation. It's done this way.

mols product = mols CuSO4.5H2O x (1 mol product/1 mol CuSO4.5H2O) = ?

mols product = mols NH3 x (1 mol product/4 mols NH3) = ?

The smaller number is the amount of product formed and the reagent responsible for that is the limiting reagent.