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NeTakaya
2 years ago
12

N2 + 3H2 - 2NH3

Chemistry
1 answer:
vladimir1956 [14]2 years ago
4 0

The number of mole of nitrogen gas, N₂, needed to produce 150 g of ammonia, NH₃ is 4.41 moles

<h3>How to determine the mole of NH₃ produced </h3>
  • Mass of NH₃ = 150 g
  • Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
  • Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 150 / 17

Mole of NH₃ = 8.82 moles

<h3>How to determine the mole of N₂ needed </h3>

Balanced equation

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

2 moles of NH₃ were produced by 1 mole of N₂.

Therefore,

8.82 moles of NH₃ will be produced by = 8.82 / 2 = 4.41 moles of N₂.

Thus, 4.41 moles of N₂ is  needed for the reaction.

Learn more about stoichiometry:

brainly.com/question/14735801

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<u>Given:</u>

Energy in joules = 650 J

<u>To determine:</u>

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Answer is 0.289nm.

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wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

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Similarly for calculating (\rho)_{ave}, we use the formula

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Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

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