What mass of calcium chloride in grams would be enough to produce 14.35 g of AgCl ? (At. mass: Ca =40 ; Ag = 108 )

Question

What mass of calcium chloride in grams would be enough to produce 14.35 g of AgCl ? (At. mass: Ca =40 ; Ag = 108 ) (A) 5.55 g (B) 8.295 g (C) 16.59 g (D) 11.19 g

Tardigrade Admin · Accepted Answer

underset111 gcacl 2+2 AgNO N 3 Ca ( NO 3)2+ underset2 × 143.5 g 2 × Agcl cacl 2 required to produce 2 × 143.5 g of Agcl =111 g Cacl 2 required to produce 14.35 g of Agcl =(111 × 14.35/2 × 143.5)=5.55 g

Q. What mass of calcium chloride in grams would be enough to produce $14.35 g$ of $A g Cl$ ? (At. mass : $C a = 40; A g = 108$ )

5.55 g

8.295 g

16.59 g

11.19 g

$111 g c a c l_{2} + 2 A g NO N_{3} C a (N O_{3})_{2} + 2 \times 143.5 g 2 \times A g c l$
$c a c l_{2}$ required to produce $2 \times 143.5 g$ of $A g c l = 111 g$
$C a c l_{2}$ required to produce $14.35 g$ of $A g c l$
$= \frac{111 \times 14.35}{2 \times 143.5} = 5.55 g$

Q. What mass of calcium chloride in grams would be enough to produce 14.35g of AgCl ? (At. mass : Ca=40;Ag=108 )