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.
The correct answer for this question is this one: " a.The solution has a volume of 25 mL "
The observation that shows a quantitative observation is when you are talking about numeric data. Just like this one, <em>The solution has a volume of 25 mL </em>
Hope this helps answer your question and have a nice day ahead.
Overuse of the same chemicals can result in the pest becoming immune to the pesticides.
Standard temperature is 273 K
Standard pressure is 1 atm
We use the ideal gas equation to find out density of nitrogen gas in g/L
Ideal gas equation:
Molar mass of 
Pressure = 1 atm
Temperature = 273 K
= 1.25 g/L
Therefore, density of nitrogen gas at STP is 1.25 g/L
