Solid calcium hudoxide reacts with phosphoric acid according to the following equation:
3 Ca(OH)2+ 2H3PO4--- Ca3(PO4)2+ 6 H2O
what in volume of 0.0220 M phosphoric acid can be neutralized by 94.651 grams of Calcium hydroxide
IF:
3moL Ca(OH)2 + 2moL H3PO4 ---> 1moL Ca3(PO4)2 + 6moL H2O
3moLx74 g/moL Ca(OH)2 + 2moLx98g/moL H3PO4 ---> 1moLx310g/moL Ca3(PO4)2 + 6moLx18g/moL H2O
222 g Ca(OH)2 + 196 g H3PO4 ---> 310 g Ca3(PO4)2 + 108 g H2O
AND IF:
222g of pure Ca(OH)2 can be neutralized 196g of pure H3PO4
THEN:
How much pure phosphoric acid can be neutralized by 94.651g pure Calcium hydroxide?
94.651g Ca(OH)2 x (196g H3PO4/222g Ca(OH)2) = 83.565g of pure H3PO4
How much 0.0220M phosphoric acid solution can be obteined wich 83.565g of pure H3PO4?
[M]= Mass/(FwxV)
V= Mass/(Fwx[M])
V= 83.565g H3PO4/(98g/moLx0.0220 moL/L)
V= 83.565g H3PO4/(98g/moLx0.0220 moL/L)
V= 38.76 L H3PO4 solution.
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