Question

1. write the balanced equation for the double displacement reaction that occurs between sodium phosphate and magnesium chloride. then determine how many grams of each product can be made from 10.00 g of magnesium chloride.

Explanation:

Step1: Write the chemical reaction

The chemical formulas are sodium phosphate ($Na_3PO_4$), magnesium chloride ($MgCl_2$). The double - displacement reaction produces sodium chloride ($NaCl$) and magnesium phosphate ($Mg_3(PO_4)_2$). The unbalanced equation is $Na_3PO_4+MgCl_2
ightarrow NaCl + Mg_3(PO_4)_2$. Balancing the equation gives $2Na_3PO_4 + 3MgCl_2=6NaCl+Mg_3(PO_4)_2$.

Step2: Calculate the molar mass of $MgCl_2$

The molar mass of $Mg$ is $24.31\ g/mol$ and of $Cl$ is $35.45\ g/mol$. So, $M_{MgCl_2}=24.31 + 2\times35.45=95.21\ g/mol$.

Step3: Calculate the number of moles of $MgCl_2$

$n_{MgCl_2}=\frac{m}{M}=\frac{10.00\ g}{95.21\ g/mol}=0.105\ mol$.

Step4: Use stoichiometry to find moles of products

From the balanced equation, the mole ratio of $MgCl_2$ to $NaCl$ is $3:6 = 1:2$, so $n_{NaCl}=2\times n_{MgCl_2}=2\times0.105\ mol = 0.21\ mol$. The molar mass of $NaCl$ is $22.99+35.45 = 58.44\ g/mol$, and $m_{NaCl}=n_{NaCl}\times M_{NaCl}=0.21\ mol\times58.44\ g/mol = 12.27\ g$.
The mole ratio of $MgCl_2$ to $Mg_3(PO_4)_2$ is $3:1$, so $n_{Mg_3(PO_4)_2}=\frac{1}{3}\times n_{MgCl_2}=\frac{1}{3}\times0.105\ mol=0.035\ mol$. The molar mass of $Mg_3(PO_4)_2$ is $3\times24.31+2\times(30.97 + 4\times16)=262.87\ g/mol$, and $m_{Mg_3(PO_4)_2}=n_{Mg_3(PO_4)_2}\times M_{Mg_3(PO_4)_2}=0.035\ mol\times262.87\ g/mol = 9.20\ g$.

Answer:

The balanced equation is $2Na_3PO_4 + 3MgCl_2=6NaCl+Mg_3(PO_4)_2$. The mass of $NaCl$ is $12.27\ g$ and the mass of $Mg_3(PO_4)_2$ is $9.20\ g$.