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<u>Answer:</u> 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.
<u>Explanation:</u>
To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.
The balanced chemical equation is:
By Stoichiometry:
2 moles of Silver nitrate reacts with 1 mole of Zinc metal
So, 6.5 moles of silver nitrate will react with = of zinc metal
The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.
Therefore, silver nitrate is the limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of silver nitrate produces 2 moles of silver metal
So, 6.5 moles of silver nitrate will produce = of silver metal.
Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles
Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.
Answer:
6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
Explanation:
We are given the chemical equation:
And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.
Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:
- The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
- The ratio between NH₃ and Cu is 2:3.
- The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)
Dimensional Analysis:
- The amount of N₂ produced:
- The amount of Cu produced:
- And the amount of H₂O produced:
In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.