Answer:
The answer to your question is:
2.- C₁₂ H₂₄ O₁₂
Explanation:
2.-
Data
CH2O
molar mass = 360.3 g/mol
Molar mass of CH2O = 12 + 2 + 16 = 30g
Divide molar mass given by molar mass obtain
x = 360.3/30
x = 12
Finally
C₁₂ H₂₄ O₁₂
Molar mass = (12 x 12) + (24 x 1) + (16 x 12) = 144 + 24 + 192 = 360 g
3.- First we need to write the complete equation of the reaction and balanced it.
Then, we need to convert the mass given to moles of each compound.
After that, we need used rule of three calculate the amount of products based on the moles of reactants given.
Finally, convert the moles to grams.
4.-
a.- It is a relation between the mass of product obtain in an experiment and the mass of a product obtain theoretically times 100.
b.-
35 g of Mg reacted with excess O2
percent yield = 90%
Actual yield = ?
Formula
Percent yield = (actual yield/theoretical yield) x 100
Equation
2Mg + O2 ⇒ 2MgO
48.62 g of Mg ----------------- 80.62 g of MgO
35g ------------------ x
x = 58 g of MgO (Theoretical yield)
Theoretical yield = 58 g of MgO
Actual yield = percent yield x theoretical yield / 100
= 90 x 58 / 100
= 52. 23 g
