Question

how many moles of each element are present in 40 g of ferric sulphate?​

Answer 1

Answer:

6.022*1023

Explanation:

by multiplying you will get answer

Answer 2

Final answer: 40 g of ferric sulphate contains $0.2$ $mol$ $Fe$, $0.3$ $mol$ $S$, and $1.2$ $mol$ $O$.

Given that: We are given $40$ $g$ of ferric sulphate.

To find: We have to find how many moles of each element are present in $40$ $g$ of ferric sulphate.

Explanation:

• Chemical formula of ferric sulphate = $Fe_{2}(SO_{4})_{3}$

• $Fe_{2}(SO_{4})_{3}$ contains $2$ $Fe$ atoms, $3$ $S$ atoms, and $12$ $O$ atoms.

• The molecular mass of $Fe$ $=56$ $g$
• The molecular mass of $S$ $=32$ $g$
• The molecular mass of $O_{2}$ $=32$ $g$

          i.e. Molecular mass of $O$ $=16$ $g$

• The molar mass of $Fe_{2}(SO_{4})_{3}$ $= (2 \times 56)+(3\times 32)+(12\times 16)$

                                                          $= 112+96+192$

                                                          $=400$ $g$

• $Number of moles of Fe_{2}(SO_{4})_{3}=\frac{Given mass of Fe_{2}(SO_{4})_{3}}{Molar mass of Fe_{2}(SO_{4})_{3}}$

           Given mass of $Fe_{2}(SO_{4})_{3}$ $= 40$ $g$

           The molar mass of $Fe_{2}(SO_{4})_{3}$ $=400$ $g$

           Number of moles of $Fe_{2}(SO_{4})_{3}$ $=\frac{40}{400}$ $=0.1$ $mol$

• $1$ $mol$ of ferric sulphate, $Fe_{2}(SO_{4})_{3}$  contains $2$ $mol$ of Fe, $2$ $mol$ of S, and $12$ $mol$ of O.

• $0.1$ $mol$ of ferric sulphate, $Fe_{2}(SO_{4})_{3}$  contain $=2\times 0.1=0.2$ $mol$ $Fe$

• $0.1$ $mol$ of ferric sulphate,$Fe_{2}(SO_{4})_{3}$  contain $=3 \times 0.1=0.3$ $mol$ $S$

• $0.1$ $mol$ of ferric sulphate, $Fe_{2}(SO_{4})_{3}$  contain $=12 \times 0.1=1.2$ $mol$ $O$

• Hence, $40$ $g$ of ferric sulphate contains $0.2$ $mol$ $Fe$, $0.3$ $mol$ $S$, and $1.2$ $mol$ $O$.

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