Phosphoric acid can be prepared from phosphorus triiodide according to the reaction:
Pl3 (s) + H2O(l) -----------> H3PO4(aq) + HI(g)
If 150 g of PI3 (MM = 411.7 g/mol) is added to 250 mL OF H2O (g/mol), d = 1.00 g/mL), Identify which will be limiting and excess reagents. (10pts) How many grams of H3PO4 (MM = 97.99 g/mol) will be produced theoretically?
Pl3 (s) + H2O(l) -----------> H3PO4(aq) + HI(g)
If 150 g of PI3 (MM = 411.7 g/mol) is added to 250 mL OF H2O (g/mol), d = 1.00 g/mL), Identify which will be limiting and excess reagents. (10pts) How many grams of H3PO4 (MM = 97.99 g/mol) will be produced theoretically?
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Answer:
PI3 + 3 H2O → H3PO3 + 3 HI
PI3 + 3 H2O → H3PO3 + 3 HI150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol),
PI3 + 3 H2O → H3PO3 + 3 HI150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol),Mole ratio = 1:3:1:3
PI3 + 3 H2O → H3PO3 + 3 HI150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol),Mole ratio = 1:3:1:3(150)(3) = 450g of H2O
PI3 + 3 H2O → H3PO3 + 3 HI150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol),Mole ratio = 1:3:1:3(150)(3) = 450g of H2OTherefore = 250/450= 0.56moles
PI3 + 3 H2O → H3PO3 + 3 HI150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol),Mole ratio = 1:3:1:3(150)(3) = 450g of H2OTherefore = 250/450= 0.56molesTherefore H3PO3 = 1:3
PI3 + 3 H2O → H3PO3 + 3 HI150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol),Mole ratio = 1:3:1:3(150)(3) = 450g of H2OTherefore = 250/450= 0.56molesTherefore H3PO3 = 1:3( 0.56 moles/3) =0.1852moles
PI3 + 3 H2O → H3PO3 + 3 HI150 grams of PI3 (MM=411.7 g/mol) is added to 250 milliliters og H2O (MM=18.01 g/mol),Mole ratio = 1:3:1:3(150)(3) = 450g of H2OTherefore = 250/450= 0.56molesTherefore H3PO3 = 1:3( 0.56 moles/3) =0.1852moles(0.1852)(81.99) =15.18grames
Explanation:
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