Respuesta :
Answer is: mass of anhydrous magnesium sulfate 17,50 grams will remain.
m(MgSO₄·7H₂O) = 36,0 g.
m(MgSO₄) = ?
Make proportion: m(MgSO₄) : m(MgSO₄·7H₂O) = M(MgSO₄) : M(MgSO₄·7H₂O).
m(MgSO₄) : 36,0 g = 120,36 g/mol : 247,47 g/mol.
m(MgSO₄) = 17,50 g.
m(MgSO₄·7H₂O) = 36,0 g.
m(MgSO₄) = ?
Make proportion: m(MgSO₄) : m(MgSO₄·7H₂O) = M(MgSO₄) : M(MgSO₄·7H₂O).
m(MgSO₄) : 36,0 g = 120,36 g/mol : 247,47 g/mol.
m(MgSO₄) = 17,50 g.
Answer: The mass of anhydrous magnesium sulfate that will remain is 17.6 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of [tex]MgSO_4.7H_2O[/tex] = 36.0 g
Molar mass of [tex]MgSO_4.7H_2O[/tex] = 246.47 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }MgSO_4.7H_2O=\frac{36.0g}{246.47g/mol}=0.146mol[/tex]
The chemical equation for the heating of [tex]MgSO_4.7H_2O[/tex] follows:
[tex]MgSO_4.7H_2O\rightarrow MgSO_4+7H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of [tex]MgSO_4.7H_2O[/tex] produces 1 mole of anhydrous magnesium sulfate.
So, 0.146 moles of [tex]MgSO_4.7H_2O[/tex] will produce = [tex]\frac{1}{1}\times 0.146=0.146mol[/tex] of anhydrous magnesium sulfate
Now, calculating the mass of anhydrous magnesium sulfate by using equation 1, we get:
Molar mass of anhydrous magnesium sulfate = 120.37 g/mol
Moles of anhydrous magnesium sulfate = 0.146 moles
Putting values in equation 1, we get:
[tex]0.146mol=\frac{\text{Mass of anhydrous magnesium sulfate}}{120.37g/mol}\\\\\text{Mass of anhydrous magnesium sulfate}=(0.146mol\times 120.37g/mol)=17.6g[/tex]
Hence, the mass of anhydrous magnesium sulfate that will remain is 17.6 grams.
