Question

For the following reaction. 16.1 grams of methane (CH) are allowed to react with 462 grams of carbon tetrachloride. CH, (g) - CC14 (9) —>2CH2C12 (9) What is the FORMULA for the limiting reagent? What is the maximum amount of dichloromethane (CH2Cl2) that can be formed? grams What amount of the excess reagent remains after the reaction is complete? grams Submit Answer R etry Entire Group 8 more group attempts remaining Previous Next

Answer 1

2cH ch oal& 91 ISh 81mal H Molan mass= Cc Molan mass CHcl: Molan madニ8r almol FRom ezuatrm Imale c neact wih Imele cel N 16) gy cH eact with Cli54) T c cely cely 16 y cH, neact wi t 154 9 16-1 9Y 入eact wih 16-1 cc, 九ezuined=1s4.96259r Zut we have Mass d is excers cactant 4629y cc amaunt dexcess 1cagent emains462- IS4.9625 ニ307-637558Y F2om ezuattom Imele ce eact with Imole cp ce, cact wih CN16) CH CMI54) with 16 8 Y *l6) 46 2 462 97 cel, cact with But we have Mass df- ch, 九で2ui7ed- 48 gr cHs Limiting only 16-1 gr cH, 1eactan t

c 1s Limiting 71eactant FSom ezuat mole cH 2moles cH ch Pro duce (2x P5) Ct Clxl) gy CH Poluce 4 Placuce 16-) Prolure 170 CH 16 gr cH KIT) clt 16-1 9 16 171.0625 Y Mass d cH c 2