What can be the chemical formula of the oxide of the element “X” if it is known that 3 liters of this gas has a mass of 5.89 g?
Let’s find the amount of substance of an unknown gas:
v (gas) = V (gas) / Vm = 3 / 22.4 (mol).
Let’s find the molar mass of the unknown gas:
M (gas) = m (gas) / v (gas) = 5.89 / (3 / 22.4) = 5.89 * 22.4 / 3 = 44 (g / mol).
This molar mass corresponds to nitric oxide (I) or laughing gas:
M (N2O) = M (N) * 2 + M (O) = 14 * 2 + 16 = 44 (g / mol), i.e. X is nitrogen.
And if dioxides are also taken into account, then carbon monoxide (IV) or carbon dioxide is also suitable:
M (CO2) = M (C) + M (O) * 2 = 12 + 16 * 2 = 44 (g / mol), i.e. X is carbon.
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