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

Use the formula stated below

$\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Volume\:in\:L}}$

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

Answer 2

Answer:

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