Consider the following reaction: 3PbCl_2+2NH4_3PO_4→Pb_3PO4_2+6NH_4Cl If 72 mmol PbCl_2 is mixed with 50 mmol of NH4_3PO_4, then amount of Pb_3PO4_2…

Question

Consider the following reaction: 3PbCl_2+2NH4_3PO_4→Pb_3PO4_2+6NH_4Cl If 72 mmol PbCl_2 is mixed with 50 mmol of NH4_3PO_4, then amount of Pb_3PO4_2 formed in mmol. (nearest integer)

MARKS Content Team · Accepted Answer

Correct answer: 24. The reaction is given as, 3PbCl_2+2NH4_3PO_4→Pb_3PO4_2+6NH_4Cl The products formed in the above reaction is Pb_3PO4_2 and NH_4Cl . According to the above stoichiometric equation, three mole of each lead chloride and two moles of NH4_3PO_4 is required to make one mole of Pb_3PO4_2 and 6 moles of NH_4Cl Here the Limiting Reagent is PbCl_2 mmol of Pb_3(PO_4)_2 formed =(mmol of PbCl_2 reacted)/3=72/3 =24 m mol

Consider the following reaction:
$3 {PbCl}_{2} + 2 {({NH}_{4})}_{3} {PO}_{4} \to {Pb}_{3} {({PO}_{4})}_{2} + 6 {NH}_{4} Cl$
If $72 mmol {PbCl}_{2}$ is mixed with $50 mmol$ of ${({NH}_{4})}_{3} {PO}_{4},$ then amount of ${Pb}_{3} {({PO}_{4})}_{2}$ formed in $mmol$ . (nearest integer)

Solution

Relevant Questions

Consider the following reaction:3PbCl2+2NH43PO4→Pb3PO42+6NH4ClIf 72 mmol PbCl2 is mixed with 50 mmol of NH43PO4, then amount of Pb3PO42 formed in mmol. (nearest integer)

Solution

Relevant Questions

Consider the following reaction:
$3 {PbCl}_{2} + 2 {({NH}_{4})}_{3} {PO}_{4} \to {Pb}_{3} {({PO}_{4})}_{2} + 6 {NH}_{4} Cl$
If $72 mmol {PbCl}_{2}$ is mixed with $50 mmol$ of ${({NH}_{4})}_{3} {PO}_{4},$ then amount of ${Pb}_{3} {({PO}_{4})}_{2}$ formed in $mmol$ . (nearest integer)