It would be <span>C21H22N2O2</span>
Answer: 48800g
Explanation:
Using the mathematical relation : Moles = Mass / Molar Mass
Moles = 488
Molar mass of CaCO3 = 40 + 12 + (16 x 3) = 100g/mol
Therefore
488 = mass / 100 = 48800g
Answer:a) 11.34 g of ethane can be formed
b) is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
1.
2.
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.378 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
moles of left = (2.10-0.378) = 1.72 moles
mass of left=
According to stoichiometry :
As 1 mole of give = 1 mole of
Thus 0.378 moles of give = of
Mass of
Thus 11.34 g of ethane is formed.
Answer:
ionic bonds in the solute are broken and new ion-dipose forces in the solution are formed
Explanation: