Question

How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2

Answer 1

6.00 moles of ScCl3 will be produced.

Step-by-step 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.

Answer 2

Moles of
`ScCl_3` = 6 moles

Step-by-step explanation:

The reaction of
`Sc` and
`Cl_2` to make
`ScCl_3` is:

`2Sc+3Cl_2`⇒
`2ScCl_3`

The above reaction shows that 2 moles of Sc can react with 3 moles of
`Cl_2` to form
`ScCl_3.`

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of
`Cl_2` =
`Moles of Sc *(3 moles of Cl_2)/(2 Moles of Sc)`

Moles of
`Cl_2` =
`10 *(3 moles of Cl_2)/(2 Moles of Sc)`

Moles of
`Cl_2` =15 moles

So 15 moles of
`Cl_2` are required to react with 10 moles of
`Sc` but we have 9 moles of
`Cl_2` , it means
`Cl_2` is limiting reactant.

`Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *(2\ Moles\ o\ fScCl_3)/(3\ Moles\ of\ Cl_2)`

`Moles\ of\ ScCl_3=9 *(2\ Moles\ of\ ScCl_3)/(3\ Moles\ of\ Cl_2)`

Moles of ScCl_3= 6 moles