30.0 g of P4010 is mixed with 75.0 g of water to form phosphoric acid: P4O10 (s)

Question

30.0 g of P4010 is mixed with 75.0 g of water to form phosphoric acid: P4O10 (s)+ 6H2O (l) -----> 4 H3PO4 (aq) A.) which one is the limiting reactant? b.) how many grams of phosphoric acid will form? 30.0 g of P4010 is mixed with 75.0 g of water to form phosphoric acid: P4O10 (s)+ 6H2O (l) -----> 4 H3PO4 (aq) A.) which one is the limiting reactant? b.) how many grams of phosphoric acid will form? 30.0 g of P4010 is mixed with 75.0 g of water to form phosphoric acid: P4O10 (s)+ 6H2O (l) -----> 4 H3PO4 (aq) A.) which one is the limiting reactant? b.) how many grams of phosphoric acid will form?

Explanation / Answer

change all to moles

mol of P4O10 = mass/MW = 30/283.886 = 0.10567 mol

mol of H2O = mass/MW = 75/18 = 4.1666 mol

ratio is 1:6

0.10567 mol of P4O10 needs = 0.10567 *6 = 0.63402

clearly, H2O is in exces

so

P410 is limiting reaction

b)

how much H3PO4 forms

1 mol of P4O10 --> 4 mol of acid

0.10567 mol of P4O10 reacts --> 0.10567 *4 = 0.42268 mol of acid

mass = mol*MW= 97.9952*0.42268 = 41.420 g of H3PO4

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