If 35.4 liters of hydrogen gas are used, how many liters of nitrogen gas will be needed for the above reaction at stp ​

11.8 L

Explanation:

Step 1: Write the balanced reaction between hydrogen and nitrogen

3 H₂(g) + N₂(g) = 2 NH₃(g)

Step 2: Calculate the moles corresponding to 35.4 L of H₂

At STP, 1 mole of H₂ occupies 22.4 L.

35.4 L × 1 mol/22.4 L = 1.58 mol

Step 3: Calculate the moles of N₂ required to react with 1.58 moles of H₂

The molar ratio of H₂ to N₂ is 3:1. The moles of N₂ required are 1/3 × 1.58 mol = 0.527 mol

Step 4: Calculate the volume occupied by 0.527 moles of N₂ at STP

At STP, 1 mole of N₂ occupies 22.4 L.

0.527 mol × 22.4 L/1 mol = 11.8 L

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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