Question

Balance the following equations:

i. H2SO4 + ….NaHCO3 → Na2SO4 + ….H2O + ……CO2
ii. H2 + I2 → ….HI
iii. …..NaOH + H2SO4 → Na2SO4 + …..H2O
iv. ……FeSO4 → Fe2O3 + SO2 + SO3
v. C3H6O2 + …..O2 → ….CO2 + …..H2O
b) Calculate the molecular masses of the following:
i. H2SO4

ii. CH4

iii. NH3

c) 56g of iron reacts with 71g of chlorine. How many grams of iron react with 35.5g of chlorine? ​

Answer 1

Hey There!

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Answer:

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Balancing the following equations:

(I)
`H_2SO_4+ 2NaHCO_3--> Na_2SO_4 + 2 H_2O + 2CO_2`

(II)
`H_2+I_2 --> 2HI`

(III)
`2NaOH+H_2SO_4--> Na_2SO_4+ 2H_2O`

(IV)
`2FeSO_4--> Fe_2O_3+SO_2+SO_3`

(V)
`2C_3H_6O_2+7O_2-->6CO_2+6H_2O`

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Molecular Masses of the following:

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(I)
`H_2SO_4`

(ANS)
`98 (grams)/(mol)`

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Solution:

Molecular mass of,

H = 1

S = 32

O = 16

Now,

There are 2 atoms of Hydrogen, 1 atoms of Sulfur and 4 atoms of Oxygen present, thus,

`H_2SO_4 = (1x2)+32+(16x4)`

`H_2SO_4 = 98 (grams)/(mol)`

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(II)
`CH_4`

(ANS)
`16 (gram)/(mol)`

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Solution:

Molecular Mass,

Carbon = 12

Hydrogen = 1

Now

There are 1 atom of Oxygen and 4 atoms of Hydrogen present, thus,

`CH_4 = 12 + (1x4)`

`CH_4 = 16(grams)/(mol)`

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(III)
`NH_3`

(ANS)
`17 (grams)/(mol)`

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Solution:

Molecular Mass,

Nitrogen = 14 grams

Hydrogen = 1

Now,

There are 1 atoms of Nitrogen and 3 atoms of Hydrogen present, thus,

`NH_3 = 14 + (1x3)`

`NH_3 = 17 (grams)/(mol)`

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How many Grams of Iron(fe):

`Fe + Cl_2 --> FeCl_2`

56 71

x 35.5

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`X = (56 x 35.5)/(71)`

Simplify the equation

X = 28 grams

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Answer:

28 grams of Iron(fe) will react with chlorine.

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Best Regards,

'Borz'