Answer:
[tex]m_{Mg(OH)_2}=60.6gMg(OH)_2[/tex]
Explanation:
Hello.
In this case, since the described chemical reaction between aqueous magnesium sulfate and sodium hydroxide is:
[tex]MgSO_4+2NaOH\rightarrow Mg(OH)_2+Na_2SO_4[/tex]
Whereas can see a 2:1 mole ratio between sodium hydroxide and magnesium sulfate whose molar masses are 120.366 g/mol and 39.997 g/mol. Next step, is to compute the yielded moles of milk of magnesia yielded by both 125 g of magnesium sulfate (1:1 mole ratio) and 115 g of sodium hydroxide (2:1 mole ratio):
[tex]n_{Mg(OH)_2}^{by\ MgSO_4}=125gMgSO_4*\frac{1molMgSO_4}{120.366gMgSO_4}*\frac{1molMg(OH)_2}{1molMgSO_4} =1.04molMg(OH)_2\\\\n_{Mg(OH)_2}^{by\ NaOH}=115gNaOH*\frac{1molNaOH}{39.997gNaOH}*\frac{1molMg(OH)_2}{2molNaOH} =1.44molMg(OH)_2\\[/tex]
Thus, since magnesium sulfate yields less moles of milk of magnesia than sodium hydroxide, we infer that MgSO4 is the limiting reactant so the yielded mass of milk of magnesia (58.320 g/mol) is:
[tex]m_{Mg(OH)_2}=1.04molMg(OH)_2*\frac{58.320gMg(OH)_2}{1molMg(OH)_2}\\\\m_{Mg(OH)_2}=60.6gMg(OH)_2[/tex]
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