30 m/hrs (add me on discord if u have it Sipstic++#1460 )
I had the same question last week and my answer was C. Try that.
That is correct! Great Job High School!
The % yield if 500 g of sulfur trioxide reacted with excess water to produce 575 g of sulfuric acid is calculated using the below formula
% yield = actual yield/ theoretical yield x100
actual yield =575 grams
to calculate theoretical yield
find the moles of SO3 used =mass/molar mass
= 500g/ 80 g/mol =6.25 moles
SO3+H2O=H2SO4
by use of mole ratio of SO3 : H2SO4 which is 1:1 the moles of H2SO4 is also= 6.25 moles
the theoretical yield of H2SO4 is therefore = moles /molar mass
= 6.25 x98= 612.5 grams
%yield is therefore= 575 g/612 g x100= 93.9 %
The problem above can be solved using M1V1=M2V2 where M1 is the concentration of the concentrated, V1 is the volume of the concentrated solution, M2 is the concentration of the Dilute Solution, V2 is the Volume of the dilute solution. Hence,
(3.0 M)(V2)=(250 mL)(1.2M)
V2 (3.0)= 300
V2= 100 mL
Therefore, you need 100 mL of 3.0 M HCl to form a 250 mL of 1.2 M HCl.
