If you produce 14.5 g of o2 in the lab from a reaction that you showed by calculation could produce a maximum of 22.0 g of o2, what is your percent yield

Percent yield is simply the ratio of the actual mass produced over the theoretical mass that can be produced. Therefore the percent yield is calculated as:

% yield = (actual mass / theoretical mass) * 100%

% yield = (14.5 / 22.0) * 100%

% yield = 65.91%

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

answer7,570,82356 cubic centimeters;