A.) How many mL of a 0.300 M lead (II) nitrate solution are needed to react with
ID: 948347 • Letter: A
Question
A.) How many mL of a 0.300 M lead (II) nitrate solution are needed to react with 50.00 mL of a 0.500 M potassium chromate solution? (THIS IS A SOLUTION STOICHIOMETRY PROBLEM.) Please show steps.
B.) A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 L volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3^-1? (I got 0.2131M, but I'm not sure. Can you please verify this for me? Thank you :))
Explanation / Answer
1)
K2CrO4 + Pb(NO3)2 ---> PbCrO4(s) + 2KNO3
M1V1 = M2V2
V1*0.3 = 50*0.5
V1 = 83.33 ml = VOLUME OF Pb(NO3)2 required
2)
No of mol of Ca(no3)2 = 20/164.088 = 0.122 mol
No of mol of naNO3 = 10/84.9947 = 0.12 mol
No of mol of Al(NO3)3 = 50/212.996 = 0.235 mol
concentration of NO3- = (0.122*2+0.12+0.235*3)/5
= 0.214 M
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.