Respuesta :
Answer:
The the empirical formula for magnesium chloride based on this experiment will be [tex]MgCl_{2}[/tex]
Explanation:
Given that,
Mass of Mg = 0.50 g
Mass of magnesium chloride found = 1.99 g
Let the formula of magnesium chloride be [tex]Mg_{x}Cl_{y}[/tex]
We know that,
Molar mass of Mg= 24 g/mol
Molar mass of magnesium chloride = (24x+35.5y) g/mol
We need to calculate the moles of Mg
Using formula of moles
[tex]Number\ of\ moles=\dfrac{mass}{molar\ mass\ of\ Mg}[/tex]
Put the value into the formula
[tex]Number\ of\ moles=\dfrac{0.50}{24}[/tex]
[tex]Number\ of\ moles=0.020\ mole[/tex]
We need to calculate the mole of magnesium chloride
Using formula of moles
[tex]Number\ of\ moles=\dfrac{mass}{molar\ mass\ of\ magnesium\ chloride}[/tex]
Put the value into the formula
[tex]Number\ of\ moles=\dfrac{1.99}{24x+35.5y}\ mole[/tex]
The reaction will be,
[tex]Mg+HCl\Rightarrow Mg_{x}Cl_{y}+H_{2}[/tex]
We need to calculate the value of x and y
Using number of moles of Mg in reactant and product
Moles of Mg atom in reactant=Moles of Mg atom in product
[tex]\dfrac{0.50}{24}=x\times\dfrac{1.99}{24x+35.5y}[/tex]
[tex]0.020(24x+35.5y)=x\times1.99[/tex]
[tex]0.48x+0.71y=1.99x[/tex]
[tex]0.71y=(1.99-0.48)x[/tex]
[tex]\dfrac{x}{y}=\dfrac{0.71}{1.99-0.48}[/tex]
[tex]\dfrac{x}{y}=0.47\approx0.5[/tex]
[tex]\dfrac{x}{y}=\dfrac{1}{2}[/tex]
Hence, The the empirical formula for magnesium chloride based on this experiment will be [tex]MgCl_{2}[/tex]
