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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)

Respuesta :

Answer : The mass of [tex]Mg_3(PO_4)_2[/tex]  produced are 5.3 grams.

Solution : Given,

Mass of [tex]MgCl_2[/tex] = 10.0 g

Mass of [tex]Na_3PO_4[/tex] = 10.0 g

Molar mass of [tex]MgCl_2[/tex] = 95.21 g/mole

Molar mass of [tex]Na_3PO_4[/tex] = 163.94 g/mole

Molar mass of [tex]Mg_3(PO_4)_2[/tex] = 262.87 g/mole

First we have to calculate the moles of [tex]MgCl_2[/tex] and [tex]Na_3PO_4[/tex].

[tex]\text{ Moles of }MgCl_2=\frac{\text{ Mass of }MgCl_2}{\text{ Molar mass of }MgCl_2}=\frac{10.0g}{95.21g/mole}=0.105moles[/tex]

[tex]\text{ Moles of }Na_3PO_4=\frac{\text{ Mass of }Na_3PO_4}{\text{ Molar mass of }Na_3PO_4}=\frac{10.0g}{163.94g/mole}=0.060moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be,

[tex]2MgCl_2(aq)+3Na_3PO_4(aq)\rightarrow 6NaCl(aq)+Mg_3(PO_4)_2(s)[/tex]

From the balanced reaction we conclude that

As, 3 mole of [tex]Na_3PO_4[/tex] react with 2 mole of [tex]MgCl_2[/tex]

So, 0.060 moles of [tex]Na_3PO_4[/tex] react with [tex]0.060\times \frac{2}{3}=0.040[/tex] moles of [tex]MgCl_2[/tex]

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

Now we have to calculate the moles of [tex]Mg_3(PO_4)_2[/tex]

From the reaction, we conclude that

As, 3 mole of [tex]Na_3PO_4[/tex] react to give 1 mole of [tex]Mg_3(PO_4)_2[/tex]

So, 0.06 moles of [tex]Na_3PO_4[/tex] react to give [tex]\frac{0.060}{3}=0.020[/tex] moles of [tex]Mg_3(PO_4)_2[/tex]

Now we have to calculate the mass of [tex]Mg_3(PO_4)_2[/tex]

[tex]\text{ Mass of }Mg_3(PO_4)_2=\text{ Moles of }Mg_3(PO_4)_2\times \text{ Molar mass of }Mg_3(PO_4)_2[/tex]

[tex]\text{ Mass of }Mg_3(PO_4)_2=(0.020moles)\times (262.87g/mole)=5.3g[/tex]

Therefore, the mass of [tex]Mg_3(PO_4)_2[/tex]  produced are 5.3 grams.

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