Answered You can buy a ready-made answer or pick a professional tutor to order an original one.
a 300 ml , 0.10 M solution of potassium sulfide was mixed with 150 ml of 0.25 m lead nitrate.a. write a balanced chemical or formula equation for the reaction.b. what mass of the precipitate is forme
a 300 ml , 0.10 M solution of potassium sulfide was mixed with 150 ml of 0.25 m lead nitrate.
a. write a balanced chemical or formula equation for the reaction.
b. what mass of the precipitate is formed?
c. calculate the concentration molarity of the nitrate ion in the solution after the reaction is complete.
- @
- 3 orders completed
- ANSWER
-
Tutor has posted answer for $10.00. See answer's preview
*** K2S **** * ** (NO3)2 *** --PbS (s) +2KNO3 ******* The ************* *** ********* *** *** ****** ** ***** of Potassium ******* is (300×01)/1000= ******** and **** ******* ** (150×025)/1000=00375moles **** ***** only 003moles ** lead ******* *** ***** ** ***** the *** moles *** since **** ratio ** *** and **** mole ***** is 1:1 for **** ******** **** *********** *** so ***** ** *** ****** *** ******** ********* **** ** *********** * ************* * 717g ***** **** *** *** ** *** atomic **** ** ** and S respectively)(c) ***** ******** ******** 00375-003 * ***** ***** ** lead ******* remain *** * mole ** **** nitrate ******** * ***** of ******* ions ** ***** *** ******** = **** ***** ** ******* ions left *** **** nitrate ions **** ****** in *** ******* **** ** ** **** **** ******** of **** were formed *** * molecule ** KNO3 ***** * of ******* ions ** *** ***** ** ******* ions **** ****** ** total ***** in ******** *** *********** 0075 ********* ************* ******** * **************** * *********