Answer:
1.428 moles
Explanation:
If 0.0714 moles of N2 gas occupies 1.25 L space,
how many moles of N2 have a volume of 25.0 L?
Assume temperature and pressure stayed constant.
we experience it 0.0714 moles: 1.25L space
x moles : 25L of space
to get the x moles, cross multiply
(0.0714 x 25)/1.25
1.785/1.25 = 1.428 moles
Rip bro but I need the point
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.
Hybridisation influences the bond length and bond enthalpy strength in organic compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp3 hybrid orbital.
