Examine the statement.
1 answer:
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The question is incomplete, the complete question is:
When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.
CaCO3→CaO + CO2
<u>Answer:</u> The % yield of the product is 87.05 %
<u>Explanation:</u>
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
......(1)
We are given:
Given mass of = 80 g
Molar mass of = 100 g/mol
Putting values in equation 1, we get:
For the given chemical reaction:
By stoichiometry of the reaction:
If 1 mole of produces 1 mole of CaO
So, 0.8 moles of will produce =
of CaO
We know, molar mass of = 56 g/mol
Putting values in above equation, we get:
The percent yield of a reaction is calculated by using an equation:
......(2)
Given values:
Actual value of the product = 39 g
Theoretical value of the product = 44.8 g
Plugging values in equation 2:
Hence, the % yield of the product is 87.05 %
Answer:
Explanation:
a. CuO+ 2HCl⇒CuCl2+ H2O
b. =
= 0,05 (mol)
⇒=
=0,05 mol
⇒= 0,05×135=6,75 (g)
c. =2×
=0,1 (mol)
⇒= 0,1×36,5= 3,65 (g)
⇒=
×100=36,5 (g)
⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=×100=
×100≈ 16,67%