M(NH4OH)=2 кг w%=10%-----------------V(NH3)-?

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вопрос

To find the volume of ammonia (NH3) produced, we first need to calculate the moles of ammonium hydroxide (NH4OH) present in 2 kg of the solution.

First, calculate the number of moles of NH4OH
Molar mass of NH4OH = 14.01 (N) + 1.01 (H) + 15.99 (O) = 30.01 g/mo
2 kg = 2000
Number of moles = 2000 g / 30.01 g/mol ≈ 66.65 moles

Since ammonium hydroxide (NH4OH) decomposes into ammonia (NH3) and water (H2O), the molar ratio is 1:1. This means that the number of moles of NH3 produced will be the same as the number of moles of NH4OH present in the solution.

Therefore, the number of moles of NH3 = 66.65 moles

Now, we need to calculate the volume of NH3 at STP (standard temperature and pressure).

1 mole of any gas at STP occupies 22.4 L of volume.

Volume of NH3 = 66.65 moles * 22.4 L/mol = 1493.76 L

Therefore, the volume of ammonia (NH3) produced is approximately 1493.76 liters.