Question
Why copper matte is put in silica lined converter?
✓
Answer
Coppermatte contains $Cu_2$ Sand $FeS$. Copper matte is put in a silica-lined converter to remove the remaining $FeO$ and FeS present in the matte as slag $(FeSiO_3).$ Also, some silica is added to the silica-lined converter. Then, a hot air blast is blown. As a result, the remaining $FeS$ and $FeO$ are converted to iron silicate $(FeSiO_3)$ and $Cu_2S$ is converted into metallic copper.
$2FeS + 3O_2 → 2FeO + 2SO_2$
$FeO + SiO_2 → FeSiO_3$
$2Cu_2S + 3O_2 → 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S → 6Cu + SO_2$
$2FeS + 3O_2 → 2FeO + 2SO_2$
$FeO + SiO_2 → FeSiO_3$
$2Cu_2S + 3O_2 → 2Cu_2O + 2SO_2$
$2Cu_2O + Cu_2S → 6Cu + SO_2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\xrightarrow{\ \ \ \text{HBr} \ \ }\text{CH}_3-\text{C}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{OH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
→Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
An aromatic compound $‘A’$ on treatment with aqueous ammonia and heating forms compound $‘B’$ which on heating with $Br_2$ and $KOH$ forms a compound $‘C’$ of molecular formula $C_6H_7N.$ Write the structures and IUPAC names of compounds $A,\ B,\ C.$
→Write Nernst equation for the following cell.
$N i (s)\left| N i ^{2+}(a q)\right|\left| A g ^{+}(a q)\right| A g _{(s)}$
→$N i (s)\left| N i ^{2+}(a q)\right|\left| A g ^{+}(a q)\right| A g _{(s)}$
How does the precipitation of colloidal smoke take place in Cottrell precipitator?
State reasons for each of the following:
→- The N–O bond in ${NO_2}^-$ is shorter than the $N–O$ bond in ${NO^-}_3.$
- $SF_6$ is kinetically an inert substance.
Identify whether the following pairs of compounds are structural or geometrical isomers:
The resistance of a conductivity cell containing $0.001\ M\ KCI$ solution at $298\ K$ is $1500\ Ω$. What is the cell constant if the conductivity of $0.001\ M\ KCI$ solution at $298\ K$ is $0.146 \times 10^{–3}\ S\ cm^{–1}?$
→Write the structures of the following:
→- $BrF_3.$
- $XeF_4.$
Explain why do transition elements shows discrepancy in the variation of atomic radii.
Benzaldehyde can be obtained from benzal chloride. Write reactions for obtaining benzalchloride and then benzaldehyde from it.