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2 years ago

Consider the hydrate FeSO4 • 7H2O.

2 answers:

7 0

The molar mass of the hydrate is 278.06
The molar mass of the anhydrous salt is 151.92
The molar mass of water in the hydrate is 126.14

mr_godi [17]2 years ago

7 0

Answer: The molar mass of the hydrate $FeSO_4$ . $7H_2O$ is 278g. The molar mass of the anhydrous salt $FeSO_4$ is 152 g. The molar mass of the water in the hydrate is $H_2O$ is 126g.

Explanation: Molar mass of Fe = 56 g

Molar mass of of S = 32 g

Molar mass of O = 16 g

Molar mass of H = 1 g

Molar mass of $FeSO_4$ • $7H_2O$ = $1\times \text{molar mass of Fe}+1\times \text{molar mass of S}+11\times \text{molar mass of O}+14\times \text{molar mass of H}$

Molar mass of $FeSO_4$ • $7H_2O$ = $1\times 56+1\times 32+11\times 16+14\times 1=278g$

Molar mass of $FeSO_4=1\times \text{molar mass of Fe}+1\times \text{molar mass of S}+4\times \text{molar mass of O}$

Molar mass of $FeSO_4=1\times 56+1\times 32+4\times 16=152g$

Molar mass of $7H_2O=14\times \text{molar mass of H}+7\times \text{molar mass of O}$

Molar mass of $7H_2O=14\times 1+7\times 16=126g$

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2 years ago

How many molecules of water are needed to completely hydrolyze a polymer that is 11 monomers long?

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2 years ago

When a known quantity of compound, at a known concentration, is added to a known volume of another compound to determine the con

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Answer:

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2 years ago

If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g)

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Answer:

The correct answer would be : 33.8 g

Explanation:

Molar mass of ammonia,

Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)

= 1*14.01 + 3*1.008 = 17.034 g/mol

mass(NH3)= 25.0 g (given)

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(25.0 g)/(17.034 g/mol)

= 1.468 mol

Now,

Molar mass of O2

= 32 g/mol

mass(O2)= 45.0 g

similar as ammonia

n (O2)=(45.0 g)/(32 g/mol)

= 1.406 mol

Balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

1.83456 mol of O2 is required for 1.46765 mol of NH3

by the calculation we have only 1.40625 mol of O2

Thus, the limiting agent will be - O2

now the Molar mass of NO,

= 1*14.01 + 1*16.0

= 30.01 g/mol (similar formula used for NH3)

Balanced equation :

mol of NO formed = (4/5)* moles of O2

= (4/5)×1.40625 (from above calculation)

= 1.125 mol

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= 1.125*30.01

= 33.8 g

Thus, the correct answer would be : 33.8 g

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