Question

For the following reaction, 13.1 grams of carbon monoxide are allowed to react with 6.09 grams of water carbon monoxide(g) + waterI) - carbon dioxide(g) + hydrogen(g) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

28 almal Cogs + H₂Oos Wrig) + H219) Weight of co= 13.1g. no. of mole of co = weight 13.1g = 0.468 moo Molar mars weight of M₂0 = 6.09 g. no. of mole of H₂O = 6.ogg = 0.3383 mwe - 18:0g med Acc to reaction can, mole co react with Imol H₂o, then, 0.468 mole to react with 0.468 mol H₂O but we have only 0.3383 mol ho. Henie O is limiting reagent The formula of limiting reagent is H₂O. mode ho prodene i mole co₂, then 0.3383 mol H₂O produse 0.3383 mod co2 mas of co₂ produced = 0.33838 44 gone = 14.88529 Hence, The maximum amount of Lo, that can be fened = 14,88529 mod no. of exces mol of L = 0.468-0.3583 = 0.1297 mol mans of excers com = 0.1297 mo o x 28 g/mol - 3.63169 Hene, amount of exen reagend (10) remain after he is complete= 3.63169