Answer is D
Balance the equation
Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
![12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}](https://tex.z-dn.net/?f=12.0molH_%7B2%7D.%5Cfrac%7B2molNH_%7B3%7D%7D%7B3molH_%7B2%7D%7D%20%3D8.00molNH_%7B3%7D)
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
![4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}](https://tex.z-dn.net/?f=4.04molN_%7B2%7D.%5Cfrac%7B2molNH_%7B3%7D%7D%7B1molN_%7B2%7D%7D%20.%5Cfrac%7B17.0gNH_%7B3%7D%7D%7B1molNH_%7B3%7D%7D%20%3D137gNH_%7B3%7D)
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
![13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}](https://tex.z-dn.net/?f=13.02gNH_%7B3%7D.%5Cfrac%7B6.00gH_%7B2%7D%7D%7B34.0gNH_%7B3%7D%7D%20%3D2.30gH_%7B2%7D)
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:
![7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}](https://tex.z-dn.net/?f=7.62%20%5Ctimes%2010%5E%7B-4%7D%20gH_%7B2%7D.%5Cfrac%7B2molNH_%7B3%7D%7D%7B6.00gH_%7B2%7D%7D%20.%5Cfrac%7B6.02%5Ctimes%2010%5E%7B23%7DmoleculesNH_%7B3%7D%20%20%7D%7B1molNH_%7B3%7D%7D%3D1.53%5Ctimes10%5E%7B20%7DmoleculesNH_%7B3%7D)
The number of Zn particles (atoms of Zn ) in the first sample, and Zn ions are going to be the same,
Also mass of Zn are going to be the same, because of electrons are too small, and would not have influence on mass.