Question

If 15.0 mL of 12.0 M H3PO4 reacts with 100.0 mL of 3.50 M of Ba(OH)2 , which substances is the limiting reactant?

2 H3PO4 + 3 Ba(OH)2 -> 6 H2O + BA3 ( PO4 )2

Answer 1

hope it's helpful to you

Answer 2

Use the formula stated below

`\boxed{\sf Molarity=(Moles\:of\:solute)/(Volume\:in\:L)}`

So

#H_3PO_4

• No of moles

`\\ \tt\Rrightarrow n=0.015(12)=0.18moles`

#Ba(OH)_2

`\\ \tt\Rrightarrow n=0.1(3.5)=0.35mol`

• Barium hydroxide is more

Hence H_3PO_4 is the limiting reagent