Featured Answer

Asked on

Theoretical Yield calculation in chemistry?

C.) When 12.0 grams of mercury (II) nitrate are decomposed in the lab, it is found that 7.56 grams of mercury (II) oxide are produced. What is the actual, theoretical, and percent yield?

BALANCED EQUATION: 2 Hg(NO3)2 ---> 2 HgO + 4 NO2 + O2

I know how to get everything but the theoretical yield! I have a quiz tomorrow and this is a problem on our review we have 3 more like it so i'd just like to see one example to help me work out the rest of them! Thanks!!

Answers (2)

Collapse
3aintscvaa profile image
Collapse
aa11937895 profile image

molar mass of mercury (II) nitrate = 324.7 g/mol

molar mass of mercury (II) oxide = 216.6 g/mol

2Hg(NO₃)₂ → 2HgO + 4NO₂ + O₂

=> 2 moles of Hg(NO₃)₂ will produce 2 moles of HgO

number of moles of 12.0 g of Hg(NO₃)₂ = 12.0g / 324.7 g/mol = 0.03696 moles

=> Theoretically 2 moles of Hg(NO₃)₂ should produce 2 moles of HgO

∴ 0.03696 moles of Hg(NO₃)₂ should produce 0.03696 moles of HgO

mass of 0.03696 moles of HgO = moles x molar mass = 0.03696(216.6) = 8.01g

Theoretical Yield = 8.01g




Enter fullscreen mode Exit fullscreen mode



Enter fullscreen mode Exit fullscreen mode



Enter fullscreen mode Exit fullscreen mode

Actually 7.56 g of HgO was produced

Actual Yield = 7.56g




Enter fullscreen mode Exit fullscreen mode



Enter fullscreen mode Exit fullscreen mode

``


% Yield = 100% (Actual Yield / Theoretical Yield)


% Yield = 100% (7.56 / 8.01)


% Yield = 94.4% 


``

Enter fullscreen mode Exit fullscreen mode



Enter fullscreen mode Exit fullscreen mode


``