Calculate the mass of Agl (fw=234.8 g/mol) that van be produced from addition of excess AgNO3 to 0.698 g sample that contains 30.6% Mgl2 (fw=278.11 g/mol)

a. 0.124 g

b. 0.240 g

c. 0.361 g

d. 0.0356 g

e. 1.07 x 10³ __g__

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Correct option is (c) 0.361 g  The detailed Solution is shown below 2 \mathrm{AgNO}_{3}+\mathrm{MgI}_{2}(\mathrm{aq}) \quad \rightarrow 2 \mathrm{AgI}\left(\Omega+\mathrm{Mg} \mathrm{NNO}_{3}\right)_{2 \text { (a) }}mislar mas of \mathrm{AgI}=234.8 \mathrm{~g} / \mathrm{nol} (given)molar nas of \frac{\left.\text { gas}_{2}\right]}{}=\mathrm{MgI}_{2}=278.11 \mathrm{glmol} (given)\begin{aligned}0.698 \mathrm{~g} \text { sample contains } & =\left(0.698 \times \frac{30.6}{100}\right) \mathrm{q} \\& =0.213588 \mathrm{gg} \mathrm{IgI}\end{aligned}from itoichiometry1 \mathrm{~mol} \mathrm{HgI}_{2} gives 2 moles \mathrm{AgI}.278.11 \mathrm{~g} \mathrm{gg}_{2} giver (2 \times 234.8 \mathrm{~g}) \mathrm{Ag} I=469.6 \mathrm{~g} \mathrm{AgI} \text {. }Hence 0.213588 \mathrm{ggg}_{2} gives =\frac{469.6}{278.11} \times 0.213585=0.36065199 \mathrm{~g} \mathrm{AgI} \text {. }option (c) 0.361 \mathrm{~g}PLEASE LIKE ! HAPPY 'ING ud83dude42 ...