The magnitude of work done (in KCal) when 650 gm of zinc is reacted with excess of HCl(aq) in an open
vessel at 300 K is
[R $=$ 2 cal $m o l^{- 1},$ At mass of Zn $=$ 65]

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Solution

The correct option is A 6
$Z n_{(S)} + 2 H C l_{(a q)} \to Z n C l_{2 (a q)} + H_{2 (g)} \to$
work done $= - Δ n \times R T$
$= - Δ n \times R T$
$= \frac{10 \times 2 \times 300}{1000}$
$= - 6 K C a l$

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Similar questions

=

^{- 1},

=

168 g

56 g / m o l

d i l . H C l

R \approx 2 c a l / m o l K

0.5 m o l

N_{2}

1.5

H_{2}

300 K

N_{2} (g) + 3 H_{2} (g) \to {2 N H}_{3} (g); Δ_{r} H_{300} = - 380 k c a l / m o l

65.38 g

300 K

Z n = 65.38

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