<span>So the oxidizing agent will receive electrons from the reducing agent and the oxidation agent will take electrons from the reducing agent.</span>
Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require = 1 mole of 
Thus 0.061 moles of will require=
of 
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 1 mole of 
Thus 0.061 moles of give = of
Mass of 
Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
Answer:
Explanation:
Num of molecules = num of moles * Avogadro's constant (6.02* 10^23)
But num of moles = reacting mass / molar mass
Molar mass of H20= 2*1 + 16 = 2+16 = 18g
Reacting mass of H20 = 0.55g
Therefore, num of moles of H20 = 0.55g/18g = 0.031 moles
Therefore, num of molecules of H20 = 0.031 * 6.02*10^23
= 1.87*10^22 molecules of H20
