Answer:
34.68g of Lithium are needed.
Explanation:
1st) It is necessary to write the balanced chemical equation:
[tex]Ca(NO_3)_2+2Li\rightarrow2Li(NO3)+Ca[/tex]
From the stoichiometry of the reaction we know that 1 mole of Calcium nitrate reacts with 2 moles of lithium.
2nd) It is necessary to use the molar mass of Ca(NO3)2 (164.09g/mol) and Li (6.94g/mol) to convert moles to grams:
• Ca(NO3)2 conversion: It is 1 mole, so the conversion is 164.09g,.
• Li conversion,:
[tex]2moles*\frac{6.94g}{1mole}=13.88g[/tex]
Now we know that 164.09g of Ca(NO3)2 react with 13.88g of Li.
3rd) Using a mathematical rule of three and with the grams of Ca(NO3)2 and Li, we can calculate the amount of lithium needed:
[tex]\begin{gathered} 164.09gCa\left(NO_3\right)_2-13.88gLi \\ 410.0Ca\left(NO_3\right)_2-x=\frac{410.0Ca\left(NO_3\right)_2*13.88gLi}{164.09gCa\left(NO_3\right)_2} \\ x=34.68gLi \end{gathered}[/tex]
Finally, 34.68g of Lithium are needed.
