A student synthesized Ni(NH3)nCl2 from NiCl2 . 6H2O using procedures described i
ID: 558240 • Letter: A
Question
A student synthesized Ni(NH3)nCl2 from NiCl2 . 6H2O using procedures described in the lab manual. The student dissolved 0.90g of the product, Ni(NH3)nCl2, in 40 mL of 1.0M HCl to make a sample for analysis. Answer the following showing all of your calculations and include units and proper significant figures.
Molar masses: Ni: 58.7 g/mol, Cl: 35.5 g/mol, NH3: 17 g/mol.
(a) The student created a standard curve in the same way we did in lab and obtained the following graph: the absorbance of his product was 0.0199. Calculate the mass percent of Ni in the product.
(b) The student used 25.00mL of 0.336M NaOH to neutralize the excess HCl in 20 mL of the sample prepared for analysis. Calculate the mass percent of NH3 in the product.
(c) Find the empirical formula for the Ni(NH3)nCl2.
Concentration of Nickel 0.12 y =0.2x + 3E-17 R2 1 0.1 4 0.08 0.06 0.04 0.02 0 0 0.1 0.2 0.3 0.4 0.5 0.6 concentration, mol/LExplanation / Answer
a) From the graph
The equation represents the straight line
y=0.2x+3.10-17
3.10-17 can be neglected as it is very less
y=0.2x
y=0.199===> 0.2x=0.199 ===> x=0.199/2= 0.0995
x= molarity of Ni2+
No of moles of Ni in 40 ml sample = molarity x vol in litres = 0.0995 x 40/1000= 0.00398
weight of Ni in the prepared 40 ml sample = molarity x At.wt x Vol in lit =0.0995 x 58.7 x 40/1000= 0.2336 gms
weight percentage of Ni in the sample = (0.2336/0.9)x 100 = 25.95%
b)No of moles in 40 mL of 1.0M HCl = molarity x vol in litres= 1 x 40/1000= 0.04
moles of NaOH used to neutralize excess HCl in 20 ml sample = 0.336 x 25/1000 = 0.0084
moles of NaOH used to neutralize excess HCl in 40 ml sample = 2 x 0.0084 = 0.0168
Stoichometrically, 1mole of HCl reacts with 1 mole of NH3
No of moles of HCl reacted with NH3 in the solution = 0.04-0.0168= 0.0232
No of moles of NH3 in the sample = 0.0232
Wt of NH3 in the sample = moles x Mol.Wt = 0.0232 x 17 = 0.3944
weight percentage of NH3 in the sample = (0.3944/0.9) x 100 = 43.82%
C) mole ratio of Ni and NH3 = 0.00398 : 0.0232 = 1:6 (approximately)
n= 6
Empirical formula = Ni(NH3)6Cl2
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