An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determined by using an indicator. Based on 20.00mL HNO3 being used to reach the endpoint, what was the mass of the Mg * (OH) in the antacid tablet? * 0.0583 g 0.583 5.83 g 58.3 g

0.0583g

Explanation:

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) > Mg(NO3)2(aq) + 2H2O(l)

From the question, number of moles of HNO3 reacted= concentration × volume

Concentration of HNO3= 0.100 M

Volume of HNO3 = 20.00mL

Number of moles of HNO3= 0.100 × 20/1000

Number of moles of HNO3 = 2×10^-3 moles

From the reaction equation;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2

But

n= m/M

Where;

n= number of moles of Mg(OH)2

m= mass of Mg(OH)2

M= molar mass of Mg(OH)2

m= n×M

m= 1×10^-3 moles × 58.3 gmol-1

m = 0.0583g

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the answer is 303.15 or 303