If 3.5 mL of 1.0 M Ni(NO3)2 is added to 6.00 mL of Na2S, what is mass of precipitate would you expect to recover based on the plot you prepared?
Chemical equation: Ni(NO3)2 + Ni2S -----> 2NaNO3 + NiS
Please correct me if i'm wrong at any part in here...
Reaction Mix 1.00 M Ni(NO3)2 1.00 M Na2S
Vol. mL # mol Vol. mL # mol mass of residue
1 1.0 .001 6.0 .006 .1023
2 2.0 .002 6.0 .006 .1814
3 3.0 .003 6.0 .006 .2751
4 4.0 .004 6.0 .006 .3524
5 5.0 .005 6.0 .006 .4541
6 6.0 .006 6.0 .006 .5444
7 7.0 .007 6.0 .006 .5502
8 8.o .008 6.0 .006 .5493
9 9.0 .009 6.0 .006 .5442
I just want to know if there is an easy way to calculate this and how, I would like to understand how to do it with the information on the table provided I dont just want an answer please.
Thanks!