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.
Moles = 15.5 g / 40 g/mol = 0.3875 mol
M = 0.3875 mol / 0.250 L = 1.55M
The normal boiling point<span> of </span>ethanol<span> is 78.4 degrees C and, at thistemperature, </span>the vapor pressure<span> is 101325 Pascals (Pa) or 760manometric units
thx hope this helped bye.</span>
To balance it, it would be N2 + 3H2 ------> 2NH3.
for c) it would be 2N2 + 6H2 -------> 4NH3
I am not fine..........................
