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What volume of oxygen at 35 degrees C and 2.15 atm is needed to produce 50.0g of nitric oxide?

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4NH3(g) 5O2(g)->4NO(g) 6H2O(g)

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  1. Moles NO = 50.0 g / 30.0 g/mol = 1.67

    Moles O2 = 1.67 x 5 / 4 = 2.09

    T = 35 + 273 =308 K

    V = nRT/p = 2.09 x 0.0821 x 308 / 2.15 = 24.6 L

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