Question

For the following reaction, 7.50 grams of methane (CH) are allowed to react with 25.1 grams of carbon tetrachloride methane (CH ) + carbon tetrachloride(g) dichloromethane (CH,CIX) What is the maximum mass of dichloromethane (CH2Cl2) that can be formed? What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? gram Submit Answer Retry Entire Group 3 more group attempts remaining

Answer 1

Solution CH4 (8) + Calu(g) → 2CM%C219) mass of CH4 = 7.50g moss of (44= 25.a Molar Mass of CH4 = 12 +42 16g/mol Molar Mass of ccl = 154 g/mol No. of moles of CH4 = mass - 7.50g - Molar mass 16g/mol C =0.4688 mol 8 No. of moles of cclu = mass 25.1g molar mass 154g/mol = 0.1623 mol From above reaction, imol of Chy reacts with Imol of ccly, theis 0.1623 mol of CCl4 reacts with 0.1623 mol CH4 therefore FORMULA of limiting reagent is celu Answer Molan mass of CH2 Ch=85 g/mol According to above reaction, Imole cely produces : 2 mol CH₂ Clz 0,1623 mole caly byoduces 2x 0.1623 mole - CH₂Cl2 = 0.3246 mole (h₂ cl Maximum mass of (H₂Cl2 z molar mass x no. of mole = 85g/mol X 0.3246 Maximum mass of ccl 2127.6 g Answer

Remaining mol of CH4 = 0.4688 mol-0.1623 mol = 0.3065 mol Moss of excess (H4= . no. of mole x Molas Mass : 0.3065 mol x 16g/mol = 4.99 Mass of excess reagent 24,9g| Answer