How many moles of sccl3 can be produced when 10.00 mol sc react with 9.00 mol cl2

6.00 moles of ScCl3 will be produced.

Explanation:

Step 1: Data given

Moles of Sc = 10.00 moles

Moles of Cl2 = 9.00 moles

Molar mass of Sc = 44.96 g/mol

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Sc + 3Cl2 → 2ScCl3

Step 3: Calculate the limiting reactant

For 2 moles Sc we need 3 moles Cl2 to produce 2 moles ScCl3

Cl2 is the limiting reactant. It will completely be consumed (9.00 moles)

Sc is the limiting reactant. There will react 2/3 * 9.00 = 6.00 moles

There will remain 10.00 - 6.00 =4.00 moles Sc

Step 4: Calculate moles ScCl3

For 3 moles Cl2 we'll have 2 moles ScCl3

For 9.00 moles we'll have 6.00 moles of ScCl3

6.00 moles of ScCl3 will be produced.

Just copying this over just in case others with similar questions turn this up in search results rather than the other, haha!
When you multiply or divide a negative by another negative, you get a positive. When it is a negative and a positive, it remains negative.
Assuming your question simplified into a equation (or number form? Been a while, haha!) is 32÷-2÷(-16÷4), it would end as a positive. Because you first divide the positive 32 by a negative 2, you get a negative. Because -16÷4 is in parenthesis, it is done because following the rest of the equation (P.E.M.D.A.S.!), it would be 32÷-2÷-4. Going from left to right, as you should, it results in a negative being divided by another negative, so you get a positive. 32÷-2÷(-16÷4) = 32÷-2÷-4 = -16÷-4 = 2!