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Answer:
moles, liter, concentration, and 6.0 mol/L
Explanation:
Since there aren't grams and just moles, the theoretical yield is one whole mole. If the molar mass is <span>142.04 g/mol, that is your answer. Hope I helped!</span>
You divided them both by 108 to get 125/32. you cant just leave it like that though. You turn it into a mixed number. So the correct answer is 3 29/32
Answer:
![V_{NH_3}=39.6L](https://tex.z-dn.net/?f=V_%7BNH_3%7D%3D39.6L)
Explanation:
Hello,
By assuming STP conditions (0°C and 1atm), we first compute the reacting moles of both hydrogen and nitrogen as shown below via the ideal gas equation:
![n_{N_2}=\frac{PV}{RT}=\frac{1atm*36.7L}{0.082 \frac{atm*L}{mol*K}*273.15K} =1.64molN_2\\n_{H_2}=\frac{PV}{RT}=\frac{1atm*59.4L}{0.082 \frac{atm*L}{mol*K}*273.15K} =2.65molH_2](https://tex.z-dn.net/?f=n_%7BN_2%7D%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B1atm%2A36.7L%7D%7B0.082%20%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A273.15K%7D%20%3D1.64molN_2%5C%5Cn_%7BH_2%7D%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B1atm%2A59.4L%7D%7B0.082%20%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A273.15K%7D%20%3D2.65molH_2)
Next, one identifies the limiting reagent by computing the moles of hydrogen that completely react with 1.64mol of nitrogen as follows:
![n_{H_2}^{reacting}=1.64molN_2*\frac{3molH_2}{1molN_2}=4.92molH_2](https://tex.z-dn.net/?f=n_%7BH_2%7D%5E%7Breacting%7D%3D1.64molN_2%2A%5Cfrac%7B3molH_2%7D%7B1molN_2%7D%3D4.92molH_2)
In such a way, as there are just 2.65 available moles of hydrogen one states that we have spare nitrogen and the hydrogen is the limiting reagent, thus, the yielded moles of ammonia are computed as:
![n_{NH_3}=2.65molH_2*\frac{2molNH_3}{3molH_2}=1.767molNH_3](https://tex.z-dn.net/?f=n_%7BNH_3%7D%3D2.65molH_2%2A%5Cfrac%7B2molNH_3%7D%7B3molH_2%7D%3D1.767molNH_3)
Finally, one computes the required volume in liters as:
![V_{NH_3}=\frac{nRT}{P}=\frac{1.767mol*0.082 \frac{atm*L}{mol*K} *273.15K}{1atm}\\V_{NH_3}=39.6L](https://tex.z-dn.net/?f=V_%7BNH_3%7D%3D%5Cfrac%7BnRT%7D%7BP%7D%3D%5Cfrac%7B1.767mol%2A0.082%20%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%20%2A273.15K%7D%7B1atm%7D%5C%5CV_%7BNH_3%7D%3D39.6L)
Best regards.