If 22.1 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over? 2 NO (g) + O₂ (g) → 2 NO₂ (g)
Added by Wesley F.
Step 1
Moles of NO = mass / molar mass = 22.1 g / (14.01 g/mol + 16.00 g/mol) = 22.1 g / 30.01 g/mol = 0.736 moles Moles of O₂ = mass / molar mass = 13.8 g / (2 * 16.00 g/mol) = 13.8 g / 32.00 g/mol = 0.431 moles Show more…
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