Question

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g of sodium phosphate is added to a solution containing 10.0g of calcium chloride

Answer 1

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of
`Na_3PO_4` = 12.00 g

Mass of
`CaCl_2` = 10.0 g

Molar mass of
`Na_3PO_4` = 164 g/mol

Molar mass of
`CaCl_2` = 111 g/mol

Molar mass of
`NaCl` = 58.5 g/mol

Molar mass of
`Ca_3(PO_4)_2` = 310 g/mol

First we have to calculate the moles of
`Na_3PO_4` and
`CaCl_2`.

`\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}`

`\text{Moles of }Na_3PO_4=(12.00g)/(164g/mol)=0.0732mol`

and,

`\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}`

`\text{Moles of }CaCl_2=(10.0g)/(111g/mol)=0.0901mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

`2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2`

From the balanced reaction we conclude that

As, 3 mole of
`CaCl_2` react with 2 mole of
`Na_3PO_4`

So, 0.0901 moles of
`CaCl_2` react with
`(2)/(3)* 0.0901=0.0601` moles of
`Na_3PO_4`

From this we conclude that,
`Na_3PO_4` is an excess reagent because the given moles are greater than the required moles and
`CaCl_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`NaCl` and
`Ca_3(PO_4)_2`

From the reaction, we conclude that

As, 3 mole of
`CaCl_2` react to give 6 mole of
`NaCl`

So, 0.0901 mole of
`CaCl_2` react to give
`(6)/(3)* 0.0901=0.1802` mole of
`NaCl`

and,

As, 3 mole of
`CaCl_2` react to give 1 mole of
`Ca_3(PO_4)_2`

So, 0.0901 mole of
`CaCl_2` react to give
`(1)/(3)* 0.0901=0.030` mole of
`Ca_3(PO_4)_2`

Now we have to calculate the mass of
`NaCl` and
`Ca_3(PO_4)_2`

`\text{ Mass of }NaCl=\text{ Moles of }NaCl* \text{ Molar mass of }NaCl`

`\text{ Mass of }NaCl=(0.1802moles)* (58.5g/mole)=10.5g`

and,

`\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2* \text{ Molar mass of }Ca_3(PO_4)_2`

`\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)* (310g/mole)=9.3g`

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.