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Sagot :
Explanation:
42.3 g of N2O5 were reacted with 9.18 g of H2O according to the following equation.
N₂O₅ + H₂O ----> 2 HNO₃
To determine which is the limiting reagent we will calculate the number of moles of HNO₃ that would be produced by each reactant. The one that produces less product is the one that is limiting the reagent.
N₂O₅:
First we can convert the 42.3 g of it into moles using its molar mass.
molar mass of N₂O₅ = 108.01 g/mol
moles of N₂O₅ = 42.3 g * 1 mol/(108.01 g)
moles of N₂O₅ = 0.392 moles
N₂O₅ + H₂O ----> 2 HNO₃
According to the reaction 1 mol of N₂O₅ will produce 2 moles of HNO₃. Then the molar ratio between them is 1 to 2. We can use this relationship to find the number of moles of HNO₃ that would be produced by 0.392 moles of N₂O₅.
1 mol of N₂O₅ : 2 moles of HNO₃ molar ratio
moles of HNO₃ = 0.392 moles of N₂O₅ * 2 moles of HNO₃/(1 mol of N₂O₅)
moles of HNO₃ = 0.784 moles
H₂O:
molar mass of H₂O = 18.02 g/mol
moles of H₂O = 9.18 g * 1 mol/(18.02 g)
moles of H₂O = 0.509 moles
1 mol of H₂O : 2 moles of HNO₃ molar ratio
moles of HNO₃ = 0.509 moles of H₂O * 2 moles of HNO₃/(1 mol of H₂O)
moles of HNO₃ = 1.018 moles
Conclusion:
When 42.3 g of N₂O₅ reacts with enough H₂O, 0.784 moles of HNO₃ are produced. When 9.18 g of H₂O reacts with enough N₂O₅, 1.018 moles of HNO₃ are produced. N₂O₅ is limiting our reaction and is the limiting reagent.
Answer: N₂O₅ is the limiting reagent.
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