Answer:
Moles of [tex]ScCl_3[/tex] = 6 moles
Explanation:
The reaction of [tex]Sc[/tex] and [tex]Cl_2[/tex] to make [tex]ScCl_3[/tex] is:
[tex]2Sc+3Cl_2[/tex]⇒[tex]2ScCl_3[/tex]
The above reaction shows that 2 moles of Sc can react with 3 moles of [tex]Cl_2[/tex] to form [tex]ScCl_3.[/tex]
Mole Ratio= 2:3
For 10 moles of Sc we need:
Moles of [tex]Cl_2[/tex] = [tex]Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}[/tex]
Moles of [tex]Cl_2[/tex] = [tex]10 *\frac{3 moles of Cl_2}{2 Moles of Sc}[/tex]
Moles of [tex]Cl_2[/tex] =15 moles
So 15 moles of [tex]Cl_2[/tex] are required to react with 10 moles of [tex]Sc[/tex] but we have 9 moles of [tex]Cl_2[/tex] , it means [tex]Cl_2[/tex] is limiting reactant.
[tex]Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}[/tex]
[tex]Moles\ of\ ScCl_3=9 *\frac{2\ Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}[/tex]
Moles of ScCl_3= 6 moles
Answer:
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.
