Recorded data- Before - CuO-porcelain boat -62-869g CaCl2-U-tube -80-483gAfter-Porcelain boat -contents -54-869g CaCl2-U-tube -89-483gReactions- Zns-2HClaq- ZnCl2aq-H2g then H2g-CuOs- H2Og-CusHow many grams of hydrogen were used in the formation of the water as product-

The correct option is A 1 Mass of #160; CaCl_2 + U tube = 80.483 gMass of #160; CaCl_2 + U tube #160; + H_2O= 89.483 gMass of wate ...

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Byju's Answer

Standard XII

Chemistry

Oxidation Number Method

Recorded data...

Question

Recorded data:
Before $:$ $C u O +$ porcelain boat $= 62.869$ g
$C a C l_{2} +$ U-tube $= 80.483$ g
After:Porcelain boat $+$ contents $= 54.869$ g
$C a C l_{2} +$ U-tube $= 89.483$ g
Reactions: $Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)$ then $H_{2} (g) + C u O (s) \to H_{2} O (g) + C u (s)$
How many grams of hydrogen were used in the formation of the water as product?

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Solution

The correct option is A $1$
Mass of $C a C l_{2} +$ U tube $= 80.483$ g
Mass of $C a C l_{2} +$ U tube $+ H_{2} O = 89.483$ g
Mass of water $= 89.483 - 80.483 = 9$ g
The molar mass of water is 18 g/mol.
9 g water corresponds to $\frac{9 g}{18 g / m o l} = 0.5$ moles.
1 mole of $H_{2}$ gives 1 mole of $H_{2} O$
0.5 moles of $H_{2}$ will give 0.5 g of $H_{2} O$
The molar mass of $H_{2}$ is 2 g/mol.
Mass of $H_{2} = 2 g / m o l \times 0.5 m o l = 1 g$
Hence, 1 gram of hydrogen were used in the formation of the water as product.

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Similar questions

Q. Recorded data:
Before

:

C u O +

porcelain boat

= 62.869

C a C l_{2} +

U-tube

= 80.483

g
After:Porcelain boat

+

contents

= 54.869

C a C l_{2} +

U-tube

= 89.483

g
Reactions:

Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)

then

H_{2} (g) + C u O (s) \to H_{2} O (g) + C u (s)

What type of reaction occurred in the porcelain boat?

Q. Recorded data:
Before

:

C u O +

porcelain boat

= 62.869

C a C l_{2} +

U-tube

= 80.483

g
After:Porcelain boat

+

contents

= 54.869

C a C l_{2} +

U-tube

= 89.483

g
Reactions:

Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)

then

H_{2} (g) + C u O (s) \to H_{2} O (g) + C u (s)

What was the reason placing tube of

C a C l_{2}

between the generator and the tube containing the porcelain boat?

Q. Recorded data:
Before

:

C u O +

porcelain boat

= 62.869

C a C l_{2} +

U-tube

= 80.483

g
After:Porcelain boat

+

contents

= 54.869

C a C l_{2} +

U-tube

= 89.483

g
Reactions:

Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)

then

H_{2} (g) + C u O (s) \to H_{2} O (g) + C u (s)

Which of the following is an observation rather than a conclusion?

Q. Recorded data:
Before

:

C u O +

porcelain boat

= 62.869 g

C a C l_{2} +

U-tube

= 80.483 g

After:Porcelain boat

+

contents

= 54.869 g

C a C l_{2} + U

-tube

= 89.483 g

Reactions:

Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)

then

H_{2} (g) + C u O (s) \to H_{2} O (g) + C u (s)

What conclusion can you draw after the completion of this experiment?

Q. From the recorded data:
Weight of U-tube

= 20.36

g
Weight of U-tube and calcium chloride before

= 39.32

g
Weight of U-tube and calcium chloride after

= 57.32

g
Weight of boat and contents(copper oxide)before

= 30.23

g
Weight of boat and contents after

= 14.23

g
Weight of boat

= 5.00

g
Determine the ratio of the mass of water formed to the mass of hydrogen used in the formation of water?

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Recorded data: Before: CuO+porcelain boat=62.869gCaCl2+U-tube=80.483gAfter:Porcelain boat+contents =54.869gCaCl2+U-tube=89.483gReactions:Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) then H2(g)+CuO(s)→H2O(g)+Cu(s)How many grams of hydrogen were used in the formation of the water as product?