Respuesta :
Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Ca(NO_3)_2=\frac{10g}{164g/mol}=0.061moles[/tex]
[tex]\text{Moles of} Li_2SO_4=\frac{10g}{110g/mol}=0.091moles[/tex]
[tex]Ca(NO_3)_2+Li_2SO_4\rightarrow 2LiNO_3+CaSO_4[/tex]
According to stoichiometry :
1 mole of [tex]Ca(NO_3)_2[/tex] require = 1 mole of [tex] Li_2SO_4[/tex]
Thus 0.061 moles of [tex]Ca(NO_3)_2[/tex] will require=[tex]\frac{1}{1}\times 0.061=0.061moles[/tex] of [tex]Li_2SO_4[/tex]
Thus [tex]Ca(NO_3)_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Li_2SO_4[/tex] is the excess reagent.
As 1 mole of [tex]Ca(NO_3)_2[/tex] give = 1 mole of [tex]CaSO_4[/tex]
Thus 0.061 moles of [tex]Ca(NO_3)_2[/tex] give =[tex]\frac{1}{1}\times 0.061=0.061moles[/tex] of [tex]CaSO_4[/tex]
Mass of [tex]CaSO_4=moles\times {\text {Molar mass}}=0.061moles\times 136g/mol=8.30g[/tex]
Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
