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Standard IX

Chemistry

Atomic Number and Mass Number

Excess of HCl...

Question

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. 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

Q. A mass of

0.65 g

of zinc reacts completely with

H C l

to produce hydrogen gas and a zinc chloride compound. The mass of the chlorine in the compound is

0.71 g

, what is the empirical formula of this compound? (At. mass of

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