Question

2 attempts left Check my work Be sure to answer all parts. Calculate the maximum numbers of moles and grams of HS that can form when 167.3 g of aluminum sulfide reacts with 143.0 g of water: Al,S3 + H,0 - Al(OH)3 + H2S (unbalanced) 3.34 14,5 133.9 XL, What mass of the excess reactant remains? 41.31 Xxcess reactant

Answer 1

Al, 33+ BHO A1201, the Molar mm i Ho = 18.01528 gmoti Molavr mom of His = 34:19 moh molar mom of Al3 = 150.158 g molt Ballmeed ean Al S3 + 6ho 2A12011) + 34 from the balenud egh it is Surved that I ond of reael with a moles of water to form 3 moles of ts. Al Sa po Imol Abg react with 6 moles of tho IP 150.158 g & Alireact with 6 X 18.01528 g of the 167.3 g of Also react with 6818 015284 167-3 dn 130.118 g 120-43 g of the .. Here uning reagent a Als Exem reagent = (143– 120.43) 8 of the = 22.57 g of the krom, Imol Al B, 3 mol of the apotued. re, from 150.158 g of Al3 , 3x34.1 g of the as protuced. 167.3 got Ahs3 32 34 17 167.3 got the 113.979 g of the 150.158

113.9798hs = 113.979g of the 34. 1 gmolt – 3.34 mol of 143. . 3.34 moles of this is prosued. 113.979 9 of 143 es produed. Exem reagent b 22.57 g of two. L a H OTA ON 210. 30 H Toplish 1. A orrels menwar rm rolo 2 1 00) INS- OH) ba fuit an RNH of the w 2 hite OU D 20 Hw210 B bis