A chemical technician prepares Ni(NH3)6Cl2 by adding 4.55 g of solid NiCl2·6H2O to 25 mL of 6.0 M aqueous NH3. What is the maximum mass of Ni(NH3)6Cl2 that will be produced?

how would I do this question?

First write out the equation and balance it.

NiCl2·6H2O + 6 NH3 ----> Ni(NH3)6Cl2 + 6 H20

Now it's basic stoichiometry.

1)Find moles of NiCl2·6H2O
2)Use the ratio find moles of Ni(NH3)6Cl2 which in this case is 1:1.
3)Convert moles of Ni(NH3)6Cl2 to grams answer the question

1)Find moles of NiCl2·6H2O

How to find moles of NiCl2·6H2O?

They give you grams, divide it by the molar mass.

What is the molar mass of NiCl2·6H2O?

For NiCl2

Ni = 58.70
Cl2 = 35.45 x 2 = 70.9

For 6H20

H = 1.008 x 12 = 12.096
O = 16.00 x 6 = 96

Add those up, the molar mass of NiCl2·6H2O is 237.696 g.

Like I said earlier, "They give you grams, divide it by the molar mass."

4.55 g x 1 mol/237.696 g = 0.01914...mol (grams cancel and you get moles left over)

2)Use the ratio find moles of Ni(NH3)

Great you have moles 0.01914...mol of NiCl2·6H2O.

Look at the formula. For every 1 mol of NiCl2·6H2O, you have 1 mol of Ni(NH3)6Cl2 .

So you have 0.01914...mol of Ni(NH3)6Cl2

3)Convert moles of Ni(NH3)6Cl2 to grams answer the question

How do you convert to grams when you have moles? Moles goes on the bottom to cancel moles and you're left with grams.

So you just need the molar mass of Ni(NH3)6Cl2. Let me just clarify cause it's hard to tell, but it's (NH_3)_6 not 6Cl2. So you have 6 N's, and 18 H's and 2 Cl's.

Ni = 58.70
N = 14.01 x 6 = 84.06
H = 1.008 x 3 = 18.144
Cl = 35.45 x 2 = 70.9

Molar mass of Ni(NH3)6Cl2 is 231.804

So you have 0.01914...mol of Ni(NH3)6Cl2. Convert to grams.

0.01914...mol x (231.804 g/1 mol) = 4.43 g

tldr

4.55 g x (1 mol/237.7 g) x (1 mol/1mol) x (231.804 g / 1 mol) = 4.437... 4.44 grams after sig figs.

The answer key says 4.43 g but I guess they rounded somewhere. I used exact numbers on the periodic table they gave us and stored the numbers. Btw, this is question 4 of CHEM 1300 2014 Version I at UoM for people to reference in the future. Good luck on your exam tomorrow!

Ooops I posted it twice and made a typo. When I said in part 3) "H = 1.008 x 3 = 19.144", I meant to say "H = 1.008 x 18 = 19.144" (18 instead of 3)

A crucial step is not stated in the solution above. You must use the limiting reactant when determining moles of the product. Very important that you compare the moles of NH3 with NiCl2 6H2O to see which one yields less moles, that will be the limiting reactant. In this case it is in fact NiCl2 6H2O

To find the maximum mass of Ni(NH3)6Cl2 that will be produced, you need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction, thus determining the maximum amount of product that can be formed.

1. Start by determining the moles of each reactant:
- Moles of NiCl2·6H2O: Since the molar mass of NiCl2·6H2O is 237.7 g/mol, you can calculate the moles by dividing the given mass (4.55 g) by the molar mass.
- Moles of NH3: Since the concentration of NH3 is given as 6.0 M, you can calculate the moles by multiplying the volume (25 mL) by the molar concentration (6.0 mol/L). Remember to convert mL to L.

2. Use the balanced chemical equation to determine the stoichiometric ratio between the reactants and the product. The balanced equation for the reaction is:

NiCl2·6H2O + 6 NH3 → Ni(NH3)6Cl2 + 6 H2O

The coefficient of NiCl2·6H2O is 1, meaning that the stoichiometric ratio between NiCl2·6H2O and Ni(NH3)6Cl2 is 1:1.

3. Compare the moles of the reactants to determine the limiting reactant. The reactant with the smaller number of moles is the limiting reactant. This is because it restricts the maximum amount of product that can be formed.

4. Once you've identified the limiting reactant, use its moles to calculate the moles of Ni(NH3)6Cl2 formed. Since the stoichiometric ratio between the limiting reactant and the product is 1:1, the moles of Ni(NH3)6Cl2 are equal to the moles of the limiting reactant.

5. Finally, calculate the maximum mass of Ni(NH3)6Cl2 formed by multiplying the moles of Ni(NH3)6Cl2 by its molar mass (which can be found in the periodic table).

Following these steps, you should be able to determine the maximum mass of Ni(NH3)6Cl2 that will be produced.