A 125 mL bubble of hot gases at 213 ∘C and 1.77 atm is emitted from an active volcano. Part A What is the final temperature, in
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Answer:
Mass of oxygen in glucose = 29.3g
Explanation:
Mass of glucose given is 55grams.
We are to find the mass of oxygen in this compound.
In the compound we have 6 atoms of oxygen.
Solution
To find the mass of oxygen in glucose, we calculate the formula mass of glucose. We now divide the formula mass of the oxygen atom with that of the glucose and multiply by the given mass to find the unkown mass.
Atomic mass of C = 12g
H = 1g
O = 16g
Formula mass of C₆H₁₂O₆ = {(12x6) + (1x12) + (16x6)} = 180
Mass of O in glucose = x 55
= x 55
= 0.53 x 55
Mass of oxygen in glucose = 29.3g
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 %