If an experiment 4g of m2ox oxide was reduced to 2.8g of the metal if the atomic mass of metal is 56 g/Mol the number of O atoms in oxide
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From mole concept 2 we get,
Moles of M * x = Moles of M* x
Hence,
moles of M = 4*2/(2*56 + 16x)
that is equals to => 2.8 / 56
Therefore,
8/(16x + 112) = 1/20
160 = 16x + 112
16x = 58
x = 58/16
Moles of O atoms => 4/(112 + 16*58/16)*58/16
No. of atoms = > 0.0852 * 6.023 x 10^23 atoms = > 0.513 x 10^23 atoms.
Moles of M * x = Moles of M* x
Hence,
moles of M = 4*2/(2*56 + 16x)
that is equals to => 2.8 / 56
Therefore,
8/(16x + 112) = 1/20
160 = 16x + 112
16x = 58
x = 58/16
Moles of O atoms => 4/(112 + 16*58/16)*58/16
No. of atoms = > 0.0852 * 6.023 x 10^23 atoms = > 0.513 x 10^23 atoms.
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