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11. For the Following Reaction: P_(2)O_(5)(s)+3H_(2)O(t)arrow 2H_(3)PO_(4)(aq) 288 G of P_(2)O_(5) and 54 G of H_(2)O Were Supplied. A)

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11. For the following reaction: P_(2)O_(5)(s)+3H_(2)O(t)arrow 2H_(3)PO_(4)(aq) 288 g of P_(2)O_(5) and 54 g of H_(2)O were supplied. a) Which is the limiting reactant?

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To determine the limiting reactant, we need to compare the amount of each reactant available to the amount required by the balanced chemical equation.The balanced chemical equation is: Given:- 288 g of - 54 g of First, we need to calculate the molar masses of and :- Molar mass of = 2(30.97) + 5(16.00) = 141.94 g/mol- Molar mass of = 2(1.01) + 16.00 = 18.02 g/molNext, we convert the masses of and to moles:- Moles of = = 2.03 moles- Moles of = = 3.00 molesNow, we compare the mole ratio of the reactants to the stoichiometric ratio in the balanced equation:- The balanced equation requires 1 mole of for every 3 moles of .We have:- 2.03 moles of - 3.00 moles of Since we have more than enough (3.00 moles) to react with 2.03 moles of , is the limiting reactant.Therefore, the limiting reactant is .