Answer to Question #85177 in General Chemistry for Victor

Question #85177
A piece of chalk (mainly calcium carbonate) is placed in 250. mL of 0.348 M HCl. All the CaCO3 reacts, releasing carbon dioxide gas, and leaving a clear solution. 35.00 mL of the solution is pipetted into another flask. 28.8 mL of 0.0509 M NaOH is required to titrate the HCl remaining in this 35.00-mL portion. What was the original mass of CaCO3 in the piece of chalk?
1
Expert's answer
2019-02-15T05:40:31-0500

Cm=n/V and n=Cm x V = 0.348 x 0.25 =0.087 mol HCl

n=0.0509 x 0.0288 = 0.0015 mol NaOH

ntotal= n1-n2=0.087 – 0.0015 =0.0855 mol HCl

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

n=0.0855:2= 0.04275 mol CaCO3 m= n x Mr= 0.04275 x 100= 4.275


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS
paypal