Excess of $H C l$ is reacted with $6.5 g$ of zinc. The hydrogen liberated is burnt with oxygen. The mass of water formed is (Atomic mass of $Z n$ = 65)

0.2 g

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0.9 g

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1.8 g

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3.6 g

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Solution

The correct option is C $1.8 g$
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$
$n_{Z n} = \frac{6.5}{65} = 0.1$
$n_{H_{2}} f o r m e d = 0.1$
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$
$n_{H_{2}} f o r m e d = 0.1$
$W_{H_{2} O} = 18 \times 0.1 = 1.8 g m$

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Similar questions

Q. Excess of

H C l

is reacted with

6.5 g

of zinc. The hydrogen liberated is burnt with oxygen. The mass of water formed is (Atomic mass of

Z n

= 65)

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place

$Z n + 2 H C l \to Z n C l_{2} + H_{2}$

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3u

Q. Calculate the volume of Hydrogen liberated when 230 gm of sodium reacts with an excess of water ar STP. (Atomic Masses of Sodium = 23 U)
Oxygen = 16U, Hydrogen = 1U
(

N a + H_{2 O} \to N a O H + H_{2})

Q. What volume of hydrogen at NTP will be liberated when 3.25 g of zinc completely dissolve in dilute HCl?

[At. mass of Zn = 65]

Q. Hydrogen gas is prepared in the laboratory by reacting dilute

H C l

with granulated zinc. Following reaction takes place:

Z n + 2 H C l \to Z n C l_{2} + H_{2}

What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with

H C l

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Excess of HCl is reacted with 6.5 g of zinc. The hydrogen liberated is burnt with oxygen. The mass of water formed is (Atomic mass of Zn = 65)

The correct option is C 1.8 gZn+2HCl→ZnCl2+H2 nZn=6.565=0.1 nH2 formed=0.1 H2+12O2→H2O nH2 formed=0.1 WH2O=18×0.1=1.8gm

Excess of $H C l$ is reacted with $6.5 g$ of zinc. The hydrogen liberated is burnt with oxygen. The mass of water formed is (Atomic mass of $Z n$ = 65)

The correct option is C $1.8 g$
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$
$n_{Z n} = \frac{6.5}{65} = 0.1$
$n_{H_{2}} f o r m e d = 0.1$
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$
$n_{H_{2}} f o r m e d = 0.1$
$W_{H_{2} O} = 18 \times 0.1 = 1.8 g m$