Question

37.8 g of Y metal is allowed to react with 415 mL of an aqueous solution of HCl (d=1.088 g/mL) that contains 18.0% HCl by mass. Y(s) + HCl(aq) → YCl3(aq) + H2(g) (balance the reaction) a. What mass of H2(g) is produced? b. What mass of excess reactant remained at the end of this reaction?

Answer 1

13:18 RBOS Yea 36.1 249, 25% f Today ca 13:18 No. Date: / / Ansa 4/s) & nullaq) - Yg (aq) + H₂(g) We will balance the reaction and all 2444) Tonce (aq) + 2Y Cl₂ laght 2 31₂ 67 ass of Иа :) 1.088 x 415 is = 45152 g mass moles of neut 457.52 12.37 moles 36.5 mass of trium, 3) 37,8 g moles of Y trium (Y) 37.8 molly 188.9 Ibu 0.425 2 moles himiting Reagent i 4.68) moles Stoichkomiltry 2mote 4 (s) produce 3 mole H2(g). ... 0.4252 moles will produce 3x0.4252 = 0,6378moles is ley. Mass of a produced >>> 2x0.6378 =) 1. 2756g ♡ оо Send Favorite Edit Delete More =

Date: mce is the fincess Reactant. Wir Imole Y8) reacts with mal ne caq. 0.4252 mole Y reacts with 3x0.4250 mole - 2.2756 molec Moles of He present Initially 12.37 moly moles of ne Reacted = 1.2756 moles no Reacted moles of ne Romaining a 12.32 -1.2256 11, 0944 moly nie. mass of ne remaining 11.0944 X 36-5 - you. gsg