calculate the mass of Potassium chlorate required to liberate 6.of oxygen at STP molar mass of potassium chlorate is 122.5 g/mol.
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question is not fully explained , is it 6.00 g or 6 .00ml/ dm³ or something else, anyways ,let us assume 6 g Apply mole concept , moles ------> grams
2kclo3 ---> 2kcl + 3o2
2(122.5)----> 3(32)
245g --------> 96 g of oxygen
if 96 gram oxygen liberated by 245 g of kclo3
so 6 gram oxygen liberated by 245/ 96 *6= 15.31 g
now if it's 6 dm³ then
same procedure
remember 1 litre = 1 dm³=10³cm³=10³ml
245 g---------> 3( 22.4) L or 67.2 dm³ of oxygen now 6.00 dm³ oxygen required ..245 /67.2 X 6= 21 . 87 gram ....
hope it works
if so mark as brainy answer
2kclo3 ---> 2kcl + 3o2
2(122.5)----> 3(32)
245g --------> 96 g of oxygen
if 96 gram oxygen liberated by 245 g of kclo3
so 6 gram oxygen liberated by 245/ 96 *6= 15.31 g
now if it's 6 dm³ then
same procedure
remember 1 litre = 1 dm³=10³cm³=10³ml
245 g---------> 3( 22.4) L or 67.2 dm³ of oxygen now 6.00 dm³ oxygen required ..245 /67.2 X 6= 21 . 87 gram ....
hope it works
if so mark as brainy answer
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