1. How many moles of K+ ions are present in 43.1 mL of a 0.621 M K3PO4 solution?
ID: 1013444 • Letter: 1
Question
1. How many moles of K+ ions are present in 43.1 mL of a 0.621 M K3PO4 solution?
2. When aqueous solutions of (NH4)2CrO4 and Ba(NO3 )2 are combined, BaCrO4 precipitates. Calculate the mass, in grams, of the BaCrO4 produced when 2.36 mL of 0.151 M Ba(NO3 )2 and 3.94 mL of 0.646 M (NH4)2CrO4 are mixed. Calculate the mass to 3 significant figures.
3. How many mL of 3.09 M HNO3 will be needed to neutralize 185 mL of 2.37 M Ca(OH)2
4. You need to prepare 330 ml of 0.59 M HNO3. You have a 12.9 M stock solution of nitric acid. How many mL of the stock solution will you need?
Explanation / Answer
1. no of moles of K3PO4 = molarity * volume in L
= 0.621*0.0431 = 0.0267moles
K3PO4 -------> 3K+ + PO43-
0.0267 moles 3*0.0267
no of moles of K+ = 0.0801 moles
2. (NH4)2CrO4 + Ba(NO3 )2 ---------->BaCrO4 + 2NH4NO3
no of moles of (NH4)2CrO4 = molarity * volume in L
= 0.646*0.00394 = 0.002545 moles
no of moles of Ba(NO3 )2 = molarity * volume in L
= 0.151*0.00236 = 0.000356 moles
limiting reageny is Ba(NO3 )2
1 mole of Ba(NO3 )2 react with (NH4)2CrO4 to gives1 moles BaCrO4
0.000356 mole of Ba(NO3 )2 react with (NH4)2CrO4 to gives0.000356 moles BaCrO4
mass of BaCrO4 = no of moles* gram molecular mass
= 0.000356*253.32 = 0.09g >>> answer
3.2 HNO3 + Ca(OH)2 -------> Ca(NO3)2 + 2H2O
HNO3 Ca(OH)2
M1 = 3.09M M2 = 2.37M
V1 = V2 = 185ml
n1 = 2 n2 = 1
M1V1/n1 = M2V2/n2
V1 = M2V2n1/n2M1
= 2.37*185*2/1*3.09 = 283.78ml >>>> answer
4. initial final
M1= 0.59M M2 = 12.9M
V1 = 330ml V2 =
M1V1 = M2V2
V2 = M1V1/M2
= 0.59*330/12.9 = 15.09 ml >>>> answer
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