Question

Consider the balanced equation for the following reaction:

2H3PO4(aq) + 3Ca(OH)2(s) → 6H2O(l) + Ca3(PO4)2(s)

How much excess reactant remains in the reaction if 7.83 moles of H3PO4 reacts with 1.53 moles of Ca(OH)2?

7.83 moles

2.61 moles

0.510 moles

6.30 moles

6.81 moles

Answer 1

Answer 2 H3PO4 (ag) + 3 Ca (OH)20 6H204) + Caz (POL)26) Given moles of H₂PO4 = 7.83 mol moles of CaCOH)₂ = 1.53 mol In order to know which reactant is in excess Let carry out calculation у Fox Ca Coи), - 2 moles of H₂PO4 7.83 moles of H₂PO4 3 - ? moles of a cou) moles of CaCOH)2 2 .. moles of Ca COH)2 required = 3_*7.83 - 11.745 mol - But we have only 1.53 mol so all caCola would be consumed. 1) For H₂ POL 2 moles of H₂PO4 → 3 moles of cacon), ? moles of H₂PO4 → 1.53moles of CaOH) moles of H₂ pou required = 2x1.53 = 1.02 mole But us 4.23 ple of H, Po4 , 80 И, Рои H₂ Poh remains = 7.83 mol-1.02 mol = 6.18 mol excess.