the balanced equation is P4O10(s) + 6H2O(l) yields 4H3PO4(l). This reaction is c

Question

the balanced equation is P4O10(s) + 6H2O(l) yields 4H3PO4(l). This reaction is carried out using 10.0 g of P4H10 and 60.0g of H20. A) Determine the limiting reagent. B) List all compounds that are present when reaction is complete. C) Determine the theoretical yield of H3PO4 based on the starting materials and D) Determine the mass, in grams, of each compound present when the reaction is complete

Explanation / Answer

H2SO4 + Pb(CH3COO)2 >> PbSO4 + 2 CH3COOH moles H2SO4 = 7.00 g / 98.08 g/mol = 0.0714 Moles Pb(CH3COO)2 = 7.00 g / 325.292 g/mol = 0.0215 ( limiting reactant) we get 0.0215 moles of PbSO4 => 0.0215 mol x 303.264 g/mol = 6.52 g we get 2 x 0.0215 = 0.0430 moles of acetic acid => 0.0430 mol x 60.05 g/mol = 2.58 g Moles H2SO4 in excess = 0.0714 - 0.0215 = 0.0499 Mass H2SO4 = 0.0499 x 98.08 = 4.89 g

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