Nitrogen dioxide (NO2) gas and liquid water (H2O) react to form aqueous nitric acid(HNO3) and nitrogen monoxide gas. Suppose you have 2.0 mol of NO2 and 7.0 mol of H20 in a reactor. Calculate the largest amount of NHO3 that could be produced. Round your answer to the nearest 0.1 mol.

1.3 moles of HNO₃ will be produced

Explanation:

Equation for the reaction:

3NO₂ + H₂O > 2HNO₃ + NO

From the equation of the reaction, 3 moles of NO₂ reacts with 1 mole of H₂O to produce 2 moles of HNO₃

Suppose there are 2 moles of NO₂  and 7.0 moles of H₂O in a reactor, the limiting reactant will be NO₂ and H₂O will be in excess since 3 moles of NO₂ reacts with every 1 mole of H₂O.

Since 3 moles of NO₂ reacts to produce  2 moles of HNO₃;

2 moles of NO₂ will react to produce 2/3 * 2 moles HNO₃ = 1.3 moles of HNO₃

Therefore, 1.3 moles of HNO₃ will be produced

6.4 cm (max) - 25 cm (min)

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