Silicon is prepared by the reduction of K₂SiF6 with Al. Write the equation for this reaction. (Hint: Can F⁻ be oxidized in this
1 answer:
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Answer:
5A₂ + 5B .................> 4A₂B + A₂ + B
Explanation:
1- For the reactants:
We can note that:
i- we have 5 moles of substance A which is diatomic. This means that we have 5A₂
ii- we have 5 moles of substance B which is monoatomic. This means that we have 5B
Based on the above, for the reactants side, we have:
5A₂ + 5B
2- For the products:
We can note that:
i- we have 4 moles of a substance formed from the combination of two moles of substance A with one mole of substance B. This means that we have 4A₂B
ii- we have one mole of substance A that is diatomic. This means that we have A₂
iii- we have one mole of substance B that is monoatomic. This means that we have B
Based on the above, for the products side, we have:
4A₂B + A₂ + B
3- Combining reactants and products parts:
Combining the reactants side (from part 1) with the products side (from part 2), we can find that the initial equation is as follows:
5A₂ + 5B .................> 4A₂B + A₂ + B
4- Balancing the equation:
Taking a look at the initial equation we reached in part 3, we will find that:
Number of moles of substance A in reactants = 2*5 = 10 moles
Number of moles of substance A in products = 2*4 + 2*! = 10 moles
SUBSTANCE A IS BALANCED
Number of moles of substance B in reactants = 5*1 = 5 moles
Number of moles of substance B in products = 4*1 + 1 = 5 moles
SUBSTANCE B IS BALANCED
Based on the above, the final balanced equation is the same as the initial equation reached in part 3:
5A₂ + 5B .................> 4A₂B + A₂ + B
Hope this helps :)
Answer:
5A₂ + 5B .................> 4A₂B + A₂ + B
Explanation:
1- For the reactants:
We can note that:
i- we have 5 moles of substance A which is diatomic. This means that we have 5A₂
ii- we have 5 moles of substance B which is monoatomic. This means that we have 5B
Based on the above, for the reactants side, we have:
5A₂ + 5B
2- For the products:
We can note that:
i- we have 4 moles of a substance formed from the combination of two moles of substance A with one mole of substance B. This means that we have 4A₂B
ii- we have one mole of substance A that is diatomic. This means that we have A₂
iii- we have one mole of substance B that is monoatomic. This means that we have B
Based on the above, for the products side, we have:
4A₂B + A₂ + B
3- Combining reactants and products parts:
Combining the reactants side (from part 1) with the products side (from part 2), we can find that the initial equation is as follows:
5A₂ + 5B .................> 4A₂B + A₂ + B
4- Balancing the equation:
Taking a look at the initial equation we reached in part 3, we will find that:
Number of moles of substance A in reactants = 2*5 = 10 moles
Number of moles of substance A in products = 2*4 + 2*! = 10 moles
SUBSTANCE A IS BALANCED
Number of moles of substance B in reactants = 5*1 = 5 moles
Number of moles of substance B in products = 4*1 + 1 = 5 moles
SUBSTANCE B IS BALANCED
Based on the above, the final balanced equation is the same as the initial equation reached in part 3:
5A₂ + 5B .................> 4A₂B + A₂ + B
Hope this helps :)