Answer to Question #144382 in General Chemistry for Zymae Frando

Question #144382

PI3 + H2O = H3PO3 + HI
Balance the equation. If 150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol), identify the limiting and excess reagents. How many grams of H3PO3 (MM=81.99 g/mol) will be theoritically produced?

Expert's answer

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

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Answer to Question #144382 in General Chemistry for Zymae Frando

PI₃ + 3 H₂O → H₃PO₃ + 3 HI

Therefore H₃PO₃ = 1:3
( 0.56 moles/3) =0.1852moles

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Answer to Question #144382 in General Chemistry for Zymae Frando

PI3 + 3 H2O → H3PO3 + 3 HI

Therefore H3PO3 = 1:3( 0.56 moles/3) =0.1852moles

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Related Questions

PI₃ + 3 H₂O → H₃PO₃ + 3 HI

Therefore H₃PO₃ = 1:3
( 0.56 moles/3) =0.1852moles