Question

The limiting reactant is the chemical substance that determines the amount of product(s) that can ultimately be formed in a reaction. During the reaction, the limiting reactant is completely consumed or used up, and therefore, causes the reaction to stop The limiting reactant can be identified through stoichiometric calculations. After comparing the results, the reactant that produces the smaller mass of product is identified as the limiting reactant. Determine the excess reactant and calculate the mass of the remaining excess reactant after 20.0 grams of Al and 10.0 grams of O react. 4 Al(s) + 30,(9) — 2 AIO (5)

Answer 1

mass of Al=20g moles of Al= rog - 0.740nol. 27 glonel mass of O2 = log moles ef on log 32 amar - 0.312 mol. Reaction :- 4 Alls) + 2 A2O₂ (s) 3% (g) 0.312 nel nifalley: 0.74 ne we need to find the limiting Reagent (2.R) - Sleps: - Initial moles of Reactants- A1 = 0.74 mel 02 ) 0.312 mel. find mole ratio of Reactants. sic s.c → storichiometric cofficients. 41) 0.74 mal = 0.185 mes. 4 01312 mol = 0.104 mal

the least value of mole ratio is sic the limiting Reagent (LR). (0.185 mel) > 10.104mal) limiting Reagent So, (2 limiting Reagents Reactant.) value of limiting Rengent (LR) = 0.104mel Op will consume first. & ha'Al will be in excess & some amount of Al will left after Reaction. Excess amount of AL = Initial male) - (sic of) xR) I of AL - TAIM IR) = (0.74 mel) – (4 x 0.104) mas &0.324 mol Excess mole of Al= 0.329 mes Excess mass of Al= (male) x (molar mass) = 10.324 ndal x 27 g ) = 8.748g Excess mass of A1 = 8.75g guy
excess mass of Al =8.75g