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shepuryov [24]
3 years ago
7

If 3.8 moles of zinc metal react with 6.5 moles of silver nitrate, how many moles of silver metal can be formed, and how many mo

les of the excess reactant will be left over when the reaction is complete? unbalanced equation: zn + agno3 → zn(no3)2 + ag be sure to show all of your work.
Chemistry
1 answer:
bulgar [2K]3 years ago
7 0

<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:

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = \frac{1}{2}\times 6.5=3.25moles 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 = \frac{2}{2}\times 6.5=6.5moles 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.

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