Indicate whether the pair of structures shown represent stereoisomers, constitutional isomers, different conformations of the sa
me compound, or the same conformation of a compound viewed from a different perspective.
Note that cis, trans isomers are an example of stereoisomers.
Note that cis, trans isomers are an example of stereoisomers.
1 answer:
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The mass of ammonia that would be produced is 312.5 g
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
3H₂(g) + N₂(g) → 2NH₃(g)
This means
3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to produce 2 moles of ammonia.
First, we will determine the number of moles of each reactant present
For Hydrogen (H₂)
Mass = 83.6 g
Molar mass = 2.016 g/mol
Using the formula
Number of moles of H₂ present =
∴ Number of moles of H₂ present = 41.468254 moles
For Nitrogen (N₂)
Mass = 257 grams
Molar mass = 28.0134 g/mol
∴ Number of moles of N₂ present =
Number of moles of N₂ present = 9.174181 moles
Since,
3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to produce 2 moles of ammonia
Then,
27.522543 moles of hydrogen gas will react with the 9.174181 moles of nitrogen gas to produce 18.348362 moles of ammonia
∴ 18.348362 moles of ammonia will be produced during the reaction
Now, for the mass of ammonia that would be produced
From the formula
Mass = Number of moles × Molar mass
Molar mass of ammonia = 17.031 g/mol
Mass of ammonia that would be produced = 18.348362 × 17.031
Mass of ammonia that would be produced = 312.49095 g
Mass of ammonia that would be produced ≅ 312.5 g
Hence, the mass of ammonia that would be produced is 312.5 g
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Answer :
The mass of excess mass of ,
are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.
Explanation : Given,
Mass of = 4.25 g
Mass of = 7.50 g
Molar mass of = 106 g/mole
Molar mass of = 170 g/mole
Molar mass of = 276 g/mole
Molar mass of = 85 g/mole
First we have to calculate the moles of and
.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 0.044 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
The excess mole of = 0.040 - 0.022 = 0.018 mole
Now we have to calculate the mass of excess mole of .
Now we have to calculate the moles of .
As, 1 moles of react to give 1 moles of
So, 0.044 moles of react to give 0.044 moles of
Now we have to calculate the mass of .
Now we have to calculate the moles of .
As, 2 moles of react to give 2 moles of
So, 0.044 moles of react to give 0.044 moles of
Now we have to calculate the mass of .