Respuesta :
Answer : The amount of lithium nitrate will be, 438.755 g
Solution : Given,
Mass of lithium sulfate = 350 g
Molar mass of lithium sulfate = 109.94 g/mole
Molar mass of lithium nitrate = 68.9 g/mole
First we have to calculate the moles of lithium sulfate.
[tex]\text{Moles of }Li_2SO_4=\frac{\text{Mass of }Li_2SO_4}{\text{Molar mass of }Li_2SO_4}=\frac{350g}{109.94g/mole}=3.184moles[/tex]
Now we have to calculate the moles of lithium nitrate.
The given balanced reaction is,
[tex]Pb(SO_4)_2+4LiNO_3\rightarrow Pb(NO_3)_4+2Li_2SO_4[/tex]
From the reaction we conclude that
2 moles of lithium sulfate produces from 4 moles of lithium nitrate
3.184 moles of lithium sulfate produces from [tex]\frac{4}{2}\times 3.184=6.368[/tex] moles of lithium nitrate
Now we have to calculate the mas of lithium nitrate.
[tex]\text{Mass of }LiNO_3=\text{Moles of }LiNO_3\times \text{Molar mass of }LiNO_3[/tex]
[tex]\text{Mass of }LiNO_3=(6.368moles)\times (68.9g/mole)=438.755g[/tex]
Therefore, the amount of lithium nitrate will be, 438.755 g
Answer: 438.84 g
Explanation:
[tex]Pb(SO_4)_2+4LiNO_3\rightarrow Pb(NO_3)_4+ 2LiSO_4[/tex]
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]{\text {Moles of lithium sulphate}=\frac{350g}{110g/mol}=3.18moles[/tex]
As adequate amount of lead (IV) sulfate is present to do the reaction, it is an excess reagent and lithium nitrate is the limiting reagent as it limits the formation of product.
2 moles of [tex]LiSO_4[/tex] are produced from 4 moles of [tex]LiNO_3[/tex]
3.18 moles of [tex]LiSO_4[/tex] are produced from [tex]=\frac{4}{2}\times 3.18=6.36moles[/tex] of [tex]LiNO_3[/tex]
[tex]\text {Mass of lithium nitrate}= {moles}\times {\text {Molar mass}}[/tex]
[tex]\text {Mass of lithium nitrate}={6.36moles}\times {69g/mol}=438.84g[/tex]
