Answer:
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
The largest amount of HNO₃ produced from the reaction between 2 mole of NO₂ and 7.0 mole of H₂O in a reactor is 1.3 mole
We'll begin by determining the limiting reactant. This can be obtained as follow:
3NO₂ + H₂O —> 2HNO₃ + NO
From the balanced equation above,
3 moles of NO₂ reacted with 1 mole of H₂O.
Therefore,
2 moles of NO₂ will react with = [tex]\frac{2}{3}\\\\[/tex] = 0.67 mole of H₂O.
From the balanced equation above,
3 moles of NO₂ reacted to produce 2 moles of HNO₃.
Therefore,
2 moles of NO₂ will react to produce = [tex]\frac{2 *2 }{3} \\\\[/tex] = 1.3 mole of HNO₃.
Thus, the largest amount of HNO₃ produced from the reaction is 1.3 mole
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