Question

A chemist dissolves 10.0 g of MgCl2 (95.21 g/mol) and 10.0 g of Na3PO4 (163.94 g/mol) into water. The product Mg3(PO4)2 (262.87 g/mol) is insoluble in water and crashes out of solution as a white solid. Assuming the reaction goes to completion, how many grams of Mg3(PO4)2 are produced? (Hint: balance the reaction below and identify the limiting reagent) MgCl2(aq) + Na3P04(aq) → NaCl(aq) + Mg3(PO4)2(s)

Answer 1

Answer : The mass of
`Mg_3(PO_4)_2` produced are 5.3 grams.

Solution : Given,

Mass of
`MgCl_2` = 10.0 g

Mass of
`Na_3PO_4` = 10.0 g

Molar mass of
`MgCl_2` = 95.21 g/mole

Molar mass of
`Na_3PO_4` = 163.94 g/mole

Molar mass of
`Mg_3(PO_4)_2` = 262.87 g/mole

First we have to calculate the moles of
`MgCl_2` and
`Na_3PO_4`.

`\text{ Moles of }MgCl_2=\frac{\text{ Mass of }MgCl_2}{\text{ Molar mass of }MgCl_2}=(10.0g)/(95.21g/mole)=0.105moles`

`\text{ Moles of }Na_3PO_4=\frac{\text{ Mass of }Na_3PO_4}{\text{ Molar mass of }Na_3PO_4}=(10.0g)/(163.94g/mole)=0.060moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

`2MgCl_2(aq)+3Na_3PO_4(aq)\rightarrow 6NaCl(aq)+Mg_3(PO_4)_2(s)`

From the balanced reaction we conclude that

As, 3 mole of
`Na_3PO_4` react with 2 mole of
`MgCl_2`

So, 0.060 moles of
`Na_3PO_4` react with
`0.060* (2)/(3)=0.040` moles of
`MgCl_2`

From this we conclude that,
`MgCl_2` is an excess reagent because the given moles are greater than the required moles and
`Na_3PO_4` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`Mg_3(PO_4)_2`

From the reaction, we conclude that

As, 3 mole of
`Na_3PO_4` react to give 1 mole of
`Mg_3(PO_4)_2`

So, 0.06 moles of
`Na_3PO_4` react to give
`(0.060)/(3)=0.020` moles of
`Mg_3(PO_4)_2`

Now we have to calculate the mass of
`Mg_3(PO_4)_2`

`\text{ Mass of }Mg_3(PO_4)_2=\text{ Moles of }Mg_3(PO_4)_2* \text{ Molar mass of }Mg_3(PO_4)_2`

`\text{ Mass of }Mg_3(PO_4)_2=(0.020moles)* (262.87g/mole)=5.3g`

Therefore, the mass of
`Mg_3(PO_4)_2` produced are 5.3 grams.