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Question

In a particular experiment, 310 g of phosphorous (P4) reacted with excess of oxygen to form P4O10 in 75% yield. Then P4O10 formed reacted with 54 g of water to give H3PO4. Amount of H3PO4 obtained (in gm) is :
(Consider only two places after decimal for calculation)

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Solution

P4+5O2P4O10
P4O10+6H2O4H3PO4

In the firest step, for 75% yield, moles of P4O10 obtained.
= moles of P4×75100
=310124×75100
=1.875 moles

Again,
P4O10+6H2O4H3PO4
Moles of water given =5418=3 moles
Actually amount of water required for 1.875 moles of P4O10 is
=1.875×6
=11.25 moles of water.
Water is limiting agent.
6 moles of H2O produced 4 moles of H3PO4
3 moles of H2O will produce
=46×3 moles of H3PO4
=2 moles of H3PO4
Molar mass of H3PO4=98 g
Therefore, amount of H3PO4 obtained = 2×98 g=196 g

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