7.66g of ethyl butyrate is produced.
The reaction is
CH₃CH₂CH₂COOH + CH₃CH₂OH ----> CH₃CH₂CH₂COOCH₂CH₃
- The molar mass of butanoic acid is 88.11g/mol
- We have 7.45g of butanoic acid
- The moles of butanoic acid we have is 7.45/88.11 = 0.0845 mol
- If the yield is 100%, 1 mole of butanoic acid gives 1 mole of ethyl butyrate
- But the reaction yield is 78%
- 1 mole of butanoic acid gives 0.78 mole of ethyl butyrate
From 0.0845 mol of butanoic acid we get 0.78 x 0.0845 = 0.66 mol of ethyl butyrate.
The molar mass of ethyl butyrate is 116.16g/mol
So 0.66 x 116.16 = 7.66g
7.66g of ethyl butyrate is produced.
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Any elements in the same periodic group as oxygen such as sulfur, as elements are grouped together through similar properties
Answer:
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Explanation:
Through the vibration of particles in a medium.
Answer:
atomic mass 112.4 and Cadmium (Cd)
Explanation:
You have 73.35 g of MO.
After the reaction the O is removed and you only have M which the mass is 64.21 g.
With that you can calculate the mass of O removed:
Mass of O = 9.14 g
Mass = AM * moles ; (AM : Atomic Mass)
9.14 = 16 * moles
moles = 9.14 / 16
moles = 0.57125
The formula of the metal oxide is MO, meaning it has 1 mole of M per mole of O. 73.35 had 0.57125 moles of O, then it also had 0.57125 moles of M, and the remaining mass of 64.21 g represents those moles
Mass = AM * moles ; (AM : Atomic Mass)
64.21 = AM * 0.57125
AM = 64.21/0.57125 = 112.4 amu
The metal with an atomic mass of 112.4 is Cadmium (Cd), therefore the metal oxide is CdO
