Question

11.3 g of acetone, (CH_3)2 CO is mixed with 20.7 g of oxygen gas (O2) and combusted to form carbon dioxide gas and water vapor. How many grams of CO2 are formed in this process?

Answer 1

Given reaction in amoun 329/mol CH₂ CO CH 4 + 40, (g) 4 300 g) + 3H20lg) Imole 4 mole 3 mole z mole. Given convent of acetone = 11:39 = 11.3 g 0.195 mole 589 mol Amount of Oz taken=20.7% = 20etare 20.78 -0.65 mole Finct we need to determine the limiting neagent ie the reagent that is completely used or prenent is leur amount mole acetone nequines - 4 mole C. - 0.195 mole s 11 = (0.195 * 4) mole 02. = 0.78 mole O2 for complete reaction. & mole og requires :) mole acetone -) 0.65 mole og will nequine= (1 x 0.65) mole autone 0.1625 mole. 650 Oz is the limiting reagent as it can reant completely 4 mole o produces- 3 mole z proc coz. - 0.65 mole O will produce = ( x 0.65) smole wz - (2 x 0.65 x44) og (malar man. 2 - 21.45 g coz.com) of Co=449