Question

I need help with number 3( for you to help me with number 3 you have to find number 1-2) “Its an assignment”

Answer 1

34.68g of Lithium are needed.

Step-by-step explanation:

1st) It is necessary to write the balanced chemical equation:

`Ca(NO_3)_2+2Li\rightarrow2Li(NO3)+Ca`

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,:

`2moles*(6.94g)/(1mole)=13.88g`

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:

`\begin{gathered} 164.09gCa\left(NO_3\right)_2-13.88gLi \\ 410.0Ca\left(NO_3\right)_2-x=(410.0Ca\left(NO_3\right)_2*13.88gLi)/(164.09gCa\left(NO_3\right)_2) \\ x=34.68gLi \end{gathered}`

Finally, 34.68g of Lithium are needed.