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Q.
What mass of calcium chloride in grams would be enough to produce $14.35 \,g $ of $AgCl $ ? (At. mass : $Ca =40 ; Ag = 108$ )
Some Basic Concepts of Chemistry
Solution:
$\underset{111 g}{cacl _2}+2 AgNO N _3 Ca \left( NO _3\right)_2+\underset{2 \times 143.5 \,g }{2 \times Agcl }$
$cacl _2$ required to produce $2 \times 143.5\, g$ of $Agcl =111\, g$
$Cacl _2$ required to produce $14.35\, g$ of $Agcl$
$=\frac{111 \times 14.35}{2 \times 143.5}=5.55 \,g$