The answer would be 371 because it has multiple complete digits
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.
Answer:
When the pressure increases to 90.0 atm , the volume of the sample is 0.01467L
Explanation:
To answer the question, we note that
P₁ = 1.00 atm
V₁ = 1.32 L
P₂ = 90 atm.
According to Boyle's law, at constant temperature, the volume of gas is inversely proportional to its pressure
That is P₁V₁ = P₂V₂
Solving the above equation for V₂ we have
that is V₂ = 
= 
or 0.01467L
