Strategy:
Because there are two reactants, this is likely to be a limiting reagent problem. The reactant that produces fewer moles is the limiting reagent. How do we convert from amount of reactant to amount of product? Perform this calculation for each reactant, then compare the moles of product, NO, formed.
Solution:
First, carry out two sep-a-rate calculations to see which of the two reactants is the limiting reagent. Starting with 6.30 g of NH₃, calculate the number of moles of NO that could be produced if all the NH₃ reacted. The conversions are
grams of NH₃ → moles of NH₃ → moles of NO
so that
[tex]moles \: of \: NO = 6.30 \: g \: NH₃ \times \frac{1 \: mol \: NH₃}{17.030 \: g \: NH₃} \times \frac{4 \: mol \: NO}{4 \: mol \: NH₃}[/tex]
[tex]moles \: of \: NO = 0.36994 \: mol \: NO[/tex]
Next, we calculate the number of moles of NO formed from 1.80 g of O₂. The conversion steps are
grams of O₂ → moles of O₂ → moles of NO
and we write
[tex]moles \: of \: NO = 1.80 \: g \: O₂ \times \frac{1 \: mol \: O₂}{31.999 \: g \: O₂} \times \frac{4 \: mol \: NO}{5 \: mol \: O₂}[/tex]
[tex]moles \: of \: NO = 0.045001 \: mol \: NO[/tex]
Therefore, O₂ is the limiting reagent because it produces a smaller amount of NO. The mass of NO formed is
[tex]0.045001 \: mol \: NO \times \frac{30.006 \: g \: NO}{1 \: mol \: NO} = \blue{1.35 \: g \: NO}[/tex]
Answer:
The answer is C.
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