Question

Close Problem Tutored Practice Problem 8.3.2 CUSTOMERS GEDE Identify limiting reactants (maximum product method) Consider the reaction of phosphorus with water and iodine 2P (5) + 6H20 (1) +312 (5) — 6HI (aq) + 2H3PO3(aq) Determine the limiting reactant in a mixture containing 127 g of P. 248 g of H2O, and 1.39x10g of ly Calculate the maximum mass (in grams) of hydroiodic acid, HI that can be produced in the reaction The limiting reactant 15 012 H20 OP Amount of HI formed

Answer 1

The gluen balanced equation id; - 2Pls) + 6HO (1) + 31, 18) GHI (aq) + 2 H₂ Pop (991 -> The coefficients de species implies the number of moles taking part in the reaction. ie Number of moles are required for the reaction and number of moles of product formed. - Slep o calculate the number 1 mees of Reactant from the green data; mass of p ginen - 1279 Molar meus pe 31 g/mole. #moles a weight Molai mess. #moles of p = 1279 14.09 moles / 3lgimole 1891 mol mass of no a 2489 Molai meurs 7 HO : 18 g/mel. # neoles 14,0 2 9482, -113.97 moes 7 mass of I ₂ = 1.39x1039 2 13909 Moler mas 1 12 = 253.80 gmor''. #moles of I2 = 1390 gmol -15.47 moles] 253.80 a moles of P required 6 moles of tho & 3 moles of Is. 4 moles 1 P will required 12 moles of Ho & 6 moles of I,. But moles of Iz present are 5.47 only.

3 Thus, I, is present in the dimiting Amont And gets used in the reaction as a whole. Thus, dimiting reagent is (22). 3 moles of Iz produces 6 moles of HI 5.47 moles of I will produce 6x 5.47 moles of HI lo.qymotes of HI This moles of Hi produced a 10.94 moles And malde mais HI , 127.91 g/mole. 5) mass A HI - ( molai mess) *(#notes) = ( 129.919 imat Ji ( 10.94 mel) 2 1401.079 - / 1.40x6037/7 HI produced, Note :- Here, in the calculation of product formation e Amount of Hi , we use Is and math not P and Mo. Because, limiting reagent is taken for such calculation. As other reactants are present in excess and dont yield product as 100% consumption.