Question

if 65 ml of 6M Mg (OH)2 and 44 ml of 3 M H3PO4 were reacted, which. is the limiting reagent?

Answer 1

3 moles of Mg(OH)2 will react with 2 moles of H3PO4

Number of moles of Mg(OH)2 = Volume of solution (in L) * Molarity (M)

=> 65/1000 * 6

=> 0.390 moles

Number of moles of H3PO4 = Volume of solution (in L) * Molarity (M)

=> 44/1000 * 3

=> 0.132 moles

since 3 moles of Mg(OH)2 requires 2 moles of H3PO4

Hence 0.390 moles will require 0.390 * 2/3 = 0.260 moles of H3PO4 but only 0.132M of H3PO4 is present in the solution

Hence H3PO4 is the limiting reagent in the reaction