Answer to Question #268660 in General Chemistry for Min min Ners

Question

Pl3 (s) + H2O(l) -----------> H3PO3(aq) + HI(g)

If 150 g of PI3 (MM = 411.7 g/mol) is added to 250 mL OF H2O (g/mol), d = 1.00 g/mL), Identify which will be limiting and excess reagents. (10pts) How many grams of H3PO3 (MM = 97.99 g/mol) will be produced theoretically?

Answer 1

PI3 + 3 H2O → H3PO3 + 3 HI

150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol),

Mole ratio = 1:3:1:3

(150)(3) = 450g of H2O

Therefore = 250/450= 0.56moles

Therefore H3PO3 = 1:3

( 0.56 moles/3) =0.1852moles

(0.1852)(81.99) =15.18grames