Question

28 gram nitrogen and 10 gram hydrogen reacted to form Nh3
find limiting reagent
the excess reagent and by how much
how much ammonia is formed

Answer 1

Answer:

here is the answer

Explanation:

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Answer 2

Answer:

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The balanced equation of the reaction is N2 + 3H2 >>>> 2NH3

The equation says that 1 mole of nitrogen gas must react with 3 moles of hydrogen gas to yield 2 moles of ammonia gas The molar masses are

• The equation says that 1 mole of nitrogen gas must react with 3 moles of hydrogen gas to yield 2 moles of ammonia gas The molar masses are28 g of N2 + 6 g of H2 >>>>>> 34 g of NH3 and the available amounts are

• The equation says that 1 mole of nitrogen gas must react with 3 moles of hydrogen gas to yield 2 moles of ammonia gas The molar masses are28 g of N2 + 6 g of H2 >>>>>> 34 g of NH3 and the available amounts are10 g of N2 + 3 g of H2 >>>>> 13 g of NH3 if they completely react

multiplying by 2 we get: 20 g of N2 + 6 g of H2 >>>>> 26 g oNH3 if they is completely react.

• multiplying by 2 we get: 20 g of N2 + 6 g of H2 >>>>> 26 g oNH3 if they is completely react.But N2 is inadequate in amount while H2 is already just enough. We need an additional 8 g of N2 to produce 34 g of ammonia gas.

So N2 or nitrogen gas is the limiting reagent. It limits the product because it is inadequate.