Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 1 (Q.No. 10)
10.
A metal oxide is reduced by heating it in a stream of hydrogen. After complete reduction, it is found that 3.15 gm of the oxide has yielded 1.05 gm of the metal. It may be inferred that the
Discussion:
8 comments Page 1 of 1.
Ananthu said:
2 years ago
What is 2.10 and from where you guys get it? Please explain me.
Jaimin Pandya LJIET said:
4 years ago
W(M) =Weight of metal.
W(O)= Weight of oxygen.
E(M)=Equvalent weight of the metal.
E(M)= [ W(M)*8]/W(O).
= [1.05*8] /2.10.
= 4.
W(O)= Weight of oxygen.
E(M)=Equvalent weight of the metal.
E(M)= [ W(M)*8]/W(O).
= [1.05*8] /2.10.
= 4.
(2)
Hariom said:
5 years ago
(1.05*8)/2.10 = 4.
Suja said:
5 years ago
@Ayush Sharma.
How 16 comes?
How 16 comes?
Ayush sharma said:
7 years ago
M(x)O + H(2) ------ (x)M + H(2)O
16+x g------------x g
1---------------------x/(16+x)
3.15 ----------------- 3.15x/(16+x)
3.15x/(16+x)=== 1.05.
After solving this x=8 g.
Equivalent weight =8/2 =4.
16+x g------------x g
1---------------------x/(16+x)
3.15 ----------------- 3.15x/(16+x)
3.15x/(16+x)=== 1.05.
After solving this x=8 g.
Equivalent weight =8/2 =4.
Sikamani said:
7 years ago
The mass of a particular substance that can combine with or displace one gram of hydrogen or eight grams of oxygen, used in expressing combining powers, especially of elements.
M Atif said:
10 years ago
Any more explanations please?
Rohit said:
1 decade ago
Equivalent weight of an element is defined as the amount of element that combines with 8 grams of oxygen.
Here, 1.05 grams of metal combines with 2.1 grams of oxygen. So 4 grams of metal combines with 8 grams of oxygen.
Hence the Equivalent weight of metal is 4.
Here, 1.05 grams of metal combines with 2.1 grams of oxygen. So 4 grams of metal combines with 8 grams of oxygen.
Hence the Equivalent weight of metal is 4.
(4)
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