Question

. How many liters of NH3, at STP, will react with 7.4g O₂?

4NH3(g) +702(g) → 4NO2(g) + 6H₂O(g)
22

Answer 1

To determine the number of liters of NH3 that will react with 7.4g of O₂ at STP, use stoichiometry to convert the mass of O₂ to moles, and then use the balanced equation to convert moles of O₂ to moles of NH3. Finally, use the ideal gas law to convert moles of NH3 to liters at STP.

Step-by-step explanation:

To determine the number of liters of NH3 that will react with 7.4g of O₂ at STP, we need to use the balanced chemical equation:

4NH3(g) + 5O2(g) → 4NO2(g) + 6H₂O(g)

We can use stoichiometry to convert the mass of O₂ to the number of moles, and then use the molar ratio from the balanced equation to convert the moles of O₂ to moles of NH3. Finally, we can use the ideal gas law to convert the moles of NH3 to liters at STP.

First, calculate the number of moles of O₂:

7.4g O₂ * (1 mole O₂ / 32g O₂) = 0.23125 moles O₂

Next, use the molar ratio from the balanced equation to convert moles of O₂ to moles of NH3:

0.23125 moles O₂ * (4 moles NH3 / 5 moles O₂) = 0.185 moles NH3

Finally, use the ideal gas law to convert moles of NH3 to liters at STP:

0.185 moles NH3 * (22.4 liters / 1 mole) = 4.144 liters NH3