Question

in a certain operation 570gram of TiCl4 is reacted with 48 gram of Mg.to give ti and mgcl2 find moles of excess reagents left

Answer 1

Answer:

mass of TiCl4 = 358g

mass of Mg reacted = 96gm

mass of Ti is actually obtained = 32gm

molecular mass of TiCl4 = 48 + 4 × 35.5 = 190g/mol

see the reaction...

TiCl_4+2Mg\rightarrow Ti + 2MgCl_2TiCl

4

+2Mg→Ti+2MgCl

2

here we see 1 mole of TiCl4 reacts with 2 mole of Mg

or, 190g of TiCl4 reacts with 48g of Mg.

or, 358g of TiCl4 reacts with 48/190 × 358 = 90.44g but here present mass of Mg is 96g.

so, TiCl4 is limiting reagent.

now, 1 mole of TiCl4 produce 1 mole of Ti

or, 190g of TiCl4 produce 48g of Ti

or, 358g of TiCl4 produces 48/190 × 358 = 90.44gm

now % yield = experimental value /theoretical value × 100

= 32/90.44 × 100

= 35.38%