Find the mass of the rsidue left when 0.5 mole of KC103 is heated strongly.
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Answered by
2
3KCLO3->KCL+3O2. This is the balanced equation.
Now,
weight of KCLO3=122.5
Here, concept of limiting reagent should be used.
3moles of kclo3 react decomposes to give 1mole of kcl(as a residue) therefore 0.5 moles of kclo3 gives?
3moles->1mole
0.5moles->?
0.5*1/3=0.16moles
Now,
weight of KCLO3=122.5
Here, concept of limiting reagent should be used.
3moles of kclo3 react decomposes to give 1mole of kcl(as a residue) therefore 0.5 moles of kclo3 gives?
3moles->1mole
0.5moles->?
0.5*1/3=0.16moles
Dame:
Thank you for your help
sure!
Answered by
1
the compound formula is incomplete . send it properly to get correct to get correct answer.
i got that. answer is3KCLO3->KCL+3O2. weight of KCLO3=122.5. 3moles of kclo3 react decomposes to give 1mole of kcl(as a residue) therefore 0.5 moles of kclo3 gives.
3moles->1mole
0.5moles.
0.5*1/3=0.16moles
3moles->1mole
0.5moles.
0.5*1/3=0.16moles
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