3.0 g of a metal oxide on reduction gives 1.40 g of metal. Number of oxygen atoms in 3.0 g of that metal oxide is (N is Avagadro’s number)
Question
3.0 g of a metal oxide on reduction gives 1.40 g of metal.
Number of oxygen atoms in 3.0 g of that metal oxide is (N is Avagadro’s number)
Solution
To solve this problem, we need to follow these steps:
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First, we need to find the mass of oxygen in the metal oxide. We can do this by subtracting the mass of the metal from the total mass of the metal oxide. So, 3.0 g (total mass of metal oxide) - 1.40 g (mass of metal) = 1.6 g (mass of oxygen).
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Next, we need to convert the mass of oxygen to moles. We know that the molar mass of oxygen is 16.00 g/mol. So, 1.6 g ÷ 16.00 g/mol = 0.1 mol.
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Finally, we need to find the number of oxygen atoms. We know that 1 mole of any substance contains Avogadro's number (N) of particles (atoms, in this case). So, 0.1 mol × N = 0.1N.
Therefore, the number of oxygen atoms in 3.0 g of the metal oxide is 0.1N.
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