Question

I need help with number 4 please look at 1,2and 3 so you can do 4

Answer 1

The mass of the first product (calcium) in grams is 100.195 grams

Step-by-step explanation

Given that

The mass of calcium nitrate is 410.0 grams

Step 1; Write the balanced equation of the reaction

`\text{ Ca\lparen NO}_3)_2\text{ }+\text{ 2Li }\rightarrow\text{ Ca}_{(s)\text{ }}\text{ }+\text{ 2LiNO}_3`

In the above reaction, 1 mole of calcium nitrate reacts with 2 moles of Li to produce 1 mole of Ca and 2 moles of LiNO3

Step 2; Find the mole of calcium nitrate using the below formula

`\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of Ca(NO3)2 is 164.1 g/mol

`\begin{gathered} \text{ mole }=\text{ }(410)/(164.1) \\ \text{ mole }=\text{ 2.50 moles} \end{gathered}`

The moles of Calcium nitrate is 2.50 moles

Step 3; Find the mole of calcium using stoichiometry ratio

Since calcium is the first product of the reaction, hence, the number of moles of calcium can be calculated below

1 mole of Ca(NO3)2 produced 1 mole of ca

Let the number of moles of calcium be x

`\begin{gathered} \text{ 1 mole Ca\lparen NO}_3)_2\text{ }\rightarrow\text{ 1 mole Ca} \\ \text{ 2.5 moles Ca\lparen NO}_3)_2\text{ }\rightarrow\text{ x mole Ca} \\ \text{ cross multiply} \\ \text{ x mole Ca }*\text{ 1 mole Ca\lparen NO}_3)_2\text{ }=\text{ 1 mole Ca}*\text{ 2.5 moles Ca\lparen NO}_3)_2 \\ \text{ Isolate x mole Ca} \\ \text{ x mole Ca }=\text{ }\frac{1\text{ mole Ca}*2.5\cancel{moles\text{ Ca\lparen NO}_3})_2}{1\cancel{mole\text{ Ca\lparen NO}_3})_2} \\ \text{ x mole Ca }=\text{ }\frac{1\text{ }*\text{ 2.5 }}{1} \\ \text{ x }=\text{ 2.5 moles} \end{gathered}`

Therefore, the number of moles of calcium is 2.5 moles

Step 4; Find the mass of Ca using the below formula

`\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}`

recall, that the molar mass of calcium is 40.078 g/mol

`\begin{gathered} \text{ 2.5 }=\text{ }\frac{\text{ mass}}{\text{ 40.087}} \\ \text{ cross multiply} \\ \text{ mass }=\text{ 2.5 }*\text{ 40.078} \\ \text{ mass }=\text{ 100.195 grams} \end{gathered}`

Therefore, the mass of the first product (calcium) in grams is 100.195 grams