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) → YCI,(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? (Atomic weight of Y=26.98 g/mol, H=1.01g/mol, CI= 35.45 g/mol)

Answer 1

The answer is given in the attachment.

present 100 375g of Y= 3700 of Y= 1:389 mol of Y. 415 mL of HCl = ({15 xl.00p)= 418:329 Helola So, Hell 18.00418-32 g= 75.298g i Hll present = 75.298, mol = 2.065 mol. (1+35-45) Y+3HU Yl3 +3%H2. Thurs, complete H₂ produchon occurs. Mass of H₂ = 12.065 265 ) mol xo (-0202)8/mol ... MH₂ =2071 g 1.389 mol – 2:065 mol = 0.700 mola of Y remaining my a 0.100226.988 Y remaining mars 1. my 3 18.8867] Versniert