Question

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

Answer 1

NH3+H2O->NH4OH

2кг=2000 г

m=w×m÷100%
m(NH4OH)=10×2000÷100=200
M(NH4OH)=14+1×4+16+1=35
n=m÷M
n(NH4OH)=200÷35г=5,71 моль
V=n×Vm
V(NH3)=5,71×22,4=127 л

Answer 2

To calculate the volume of NH3 produced from 2 kg of NH4OH with a concentration of 10%, we first need to write the balanced chemical equation for the reaction:

NH4OH → NH3 + H2O

From the equation, we can see that 1 mole of NH4OH produces 1 mole of NH3. The molar mass of NH4OH is 35.05 g/mol (14.01 g/mol for nitrogen + 4 x 1.01 g/mol for hydrogen + 16.00 g/mol for oxygen).

First, let's calculate the number of moles of NH4OH:

2 kg x 1000 g/kg / 35.05 g/mol = 57.05 mol NH4OH

Since the concentration is 10%, we know that there is 10% NH4OH in the solution and the rest is water. Therefore, the total volume of the solution is:

Vtotal = 2 kg / 0.1 = 20 kg

Now we can calculate the mass of water in the solution:

mH2O = Vtotal - M(NH4OH) = 20 kg - 2 kg = 18 kg

The number of moles of NH3 produced will be equal to the number of moles of NH4OH used:

nNH3 = nNH4OH = 57.05 mol

Finally, we can calculate the volume of NH3 produced using the ideal gas law:

PV = nRT

Assuming standard temperature and pressure (STP) of 273 K and 1 atm:

VNH3 = nNH3 * RT/P = 57.05 mol * 0.0821 L·atm/mol·K * 273 K / 1 atm = 127.3 L

Therefore, the volume of NH3 produced is 127.3 L.