A serial dilution is the stepwise dilution of a substance in solution. Usually the dilution factor at each step is constant, resulting in a geometric progression of the concentration in a logarithmic fashion.
Answer: 292.54g of Ag
Explanation:
Cu + 2AgNO3 →Cu(NO3)2 + 2 Ag
mass conc. Of Ag = n x molar Mass
Mass conc. Of Ag = 2 x 108 = 216g
From the equation,
63.5g of Cu produced 216g of Ag
Therefore, 86g of Cu will produce Xg of Ag. i.e
Xg of Ag = (86 x 216)/63.5 = 292.54g
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%
Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.
Hey there!:
Molar mass:
H2 = 2.01 g/mol ; H2O = 18.01
Given the reaction:
2 H2 + O2 = 2 H2O
2 * (2.01 ) g H2 ------------- 2 * ( 18.01 ) g H2O
mass H2 --------------------- 1.80 g H2O
mass H2 = 1.80 * 2 * 2.01 / 2* 18.01
mass H2 = 7.236 / 36.02
mass H2 = 0.2008 g
Hope that helps!
