A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
What volume of 2.25M NaCl solution is required to react completely with 1.00L of 0.750M Pb(NO3) solution?
Pb(NO3)2(aq)+2NaCl(aq)=PbCl2(s)+2NaNO3(aq)
anonymous
 3 years ago
What volume of 2.25M NaCl solution is required to react completely with 1.00L of 0.750M Pb(NO3) solution? Pb(NO3)2(aq)+2NaCl(aq)=PbCl2(s)+2NaNO3(aq)

This Question is Closed

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.1You need to find the volume of NaCl, based from formula \[[M]=\frac{mol}{V}\]

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.1They have given some values for Pb(NO3) right, so using that formula, \[[M]=\frac{mol}{V}\] mol=MV n(Pb(NO3)) = (0.750)(1.00) = 0.750moles of Pb(NO3) According to the equation, Pb(NO3)2(aq)+2NaCl(aq)=PbCl2(s)+2NaNO3(aq) 1 Pb(NO3)2 is equivalent to 2 NaCl, so 0.750 times 2 = 1.5 moles of NaCl then use back the equation \[[M]=\frac{mol}{V}\] \[V=\frac{mol}{[M]}\] \[V=\frac{1.5}{2.25}=0.667 m^3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok thank you so much! You just cleared up the M to mol to V formula for me. I was getting stumped using M to get to other values. All I knew was that M=mol/L Thank you for your help.This was the last practice problem I couldn't figure out. I might actually be ready for this quiz today now!

VincentLyon.Fr
 3 years ago
Best ResponseYou've already chosen the best response.00.667 m3 is a lot!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.