0.02 moles of pure MnO2 is heated strongly with concentrated HCL calculate (a) mass of MnO2 used (b) moles of salt formed (c) mass of salt formed (d) moles of chlorine gas formed (e) mass of chlorine gas formed (f) volume of chlorine gas formed at STP (g) moles of acid required (h) mass of acid required.

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Solution

$M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2} a) M a s s o f M n O_{2} u s e d = M o l e \times M o l e c u l a r m a s s = 0.02 \times 86.94 = 1.74 g b) 1 m o l e o f M n O_{2} p r o d u c e s 1 m o l e o f s a l t M n C l_{2} 0.02 m o l e o f M n O_{2} p r o d u c e s 0.02 m o l e o f s a l t M n C l_{2} c) M a s s o f s a l t = M o l e \times M o l e c u l a r m a s s = 0.02 \times 125.84 = 2.52 g d) 1 m o l e o f M n O_{2} p r o d u c e s 1 m o l e o f C h l o r i n e g a s 0.02 m o l e o f M n O_{2} p r o d u c e s 0.02 m o l e o f C h l o r i n e g a s e) M a s s o f C h l o r i n e g a s = M o l e \times M o l e c u l a r m a s s = 0.02 \times 71 = 1.42 g f) 1 M o l e o f C h l o r i n e g a s o c c u p i e s 22.4 L a t S T P 0.02 m o l e o f C h l o r i n e g a s o c c u p i e s 22.4 \times 0.02 = 0.448 L o f C l_{2} g a s g) 1 M o l e o f M n O_{2} n e e d s 4 m o l e s o f H C l 0.02 m o l e o f M n O_{2} n e e d s = \frac{4}{1} \times 0.02 = 0.08 m o l e s o f H C l h) M a s s o f H C l = M o l e \times M o l e c u l a r m a s s = 0.08 \times 36.5 = 2.92 g$

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