Based on the balanced equation calculate the number of excess reagent units rema

Question

Based on the balanced equation

calculate the number of excess reagent units remaining when 95CH3CCH molecules and 396 O2 moleculesreact?

Based on the balanced equation

CH3CCH + 4O2 3CO2 + 2H2O

calculate the number of excess reagent units remaining when 95CH3CCH molecules and 396 O2 moleculesreact?

Molar Mass (g/mol) CH3CCH 40.065 O2 32.000 CO2 44.010 H2O 18.015 Avogadro's No. 6.022×1023

Explanation / Answer

n(CH3CCH) = N/(6.022 x 1023) n(CH3CCH) = 95/(6.022 x 1023) = 1.578 x10-22 mol n(O2) = N/(6.022 x 1023) n(O2) = 396/(6.022 x 1023) = 6.576 x10-22 mol Actual ratio = n(CH3CCH) / n(O2) = (1.578 x10-22 ) / (6.576 x 10-22) = 0.23996 Stoichiometry ratio = 1/4 = 0.25 SR > AR So, CH3CCH is the limiting reagent n(O2) remaining = 6.576 x 10-22 - 4 x 1.578 x10-22 = 2.66 x 10-23mol n(O2) = N / (6.022 x 1023) N = 2.66 x 10-23 x 6.022 x 1023 = 16units of O2 remaining Hope this helps!

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