Question

How many liters of a 0.75 M solution of calcium nitrate will be required to react with 148 g of sodium carbonate?

Answer 1

1.87L of the 0.75M calcium nitrate solution will be required.

Step-by-step explanation:

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

`Ca(NO_3)_2+Na_2CO_3\rightarrow CaCO_3+2Na(NO_3)`

Now we know that 1 mole of calcium nitrate (Ca(NO3)2) reacts with 1 mole of sodium carbonate (Na2CO3), to produce 1 mole of calcium carbonate (CaCO3) and 2 moles of sodium nitrate (Na(NO3)).

2nd) With the stoichiometry of the chemical reaction and the molar mass of the compounds, we can calculate the grams of calcium nitrate that are needed to react with 148g of sodium carbonate:

- Calcium nitrate molar mass: 164.1 g/mol

- Sodium carbonate molar mass: 106 g/mol

`\begin{gathered} 106gSodiumCarbonate-164.1gCalciumNitrate \\ 148gSodiumCarbonate-x=(148gSodiumCarbonat*164.1gCalciumNitrate)/(106gSodiumCarbonate) \\ x=229.12gCalciumNitrate \end{gathered}`

Now we know that 148g of sodium carbonate will need 229.12g of calcium nitrate to react properly.

Using the molar mass of calcium nitrate, we can convert the 229.12g to moles:

`229.12g*(1mol)/(164.1g)=1.4mol`

So, 1.4 moles of calcium nitrate will react.

3rd) We know that the molarity of the calcium nitrate solution is 0.75M, that means that there are 0.75 moles of calcium nitrate in 1L of solution.

Using a mathematical rule of three we can calculate the liters of this 0.75M solution of calcium nitrate are needed to react:

`\begin{gathered} 0.75moles-1L \\ 1.4moles-x=(1.4moles*1L)/(0.75moles) \\ x=1.87L \end{gathered}`

So, 1.87L of the 0.75M calcium nitrate solution will be required.