How much of lime (CaO) can be obtained by the calculations of 200gr of lime stone (Ca=40gr,O=16gr)
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Answer:
CaCO
3
→CaO+CO
2
moles of CaCO
3
in 200 kg = 1000 moles
moles of CaO produced = 1000 moles
weight of CaO produced = 1000* 56 = 56000g
limestone is 90% pure
therefore the amount of CaO produced = 56000∗
100
93
= 52080g or 52.08 kg.
we have,
KClO
3
→KCl+3/2O
2
moles of O
2
in 48 g =
32
48
= 1.5
that means 1 mole of KClO
3
is dissociating.
1 mole of KClO
3
= 122.5 g
so, amount of impure KClO
3
= 122.5∗
75
100
= 163.3 g
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