Calculate how many g of ammonia will be produced from 9.35 g of nitrogen gas and excess hydrogen using the following equation: 2 nh(g) 3 h2(g) n(g) 4 answer

The mass of ammonia produced in the reaction is 11.36 g

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of nitrogen gas = 9.35 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

The chemical reaction for the formation of ammonia from hydrogen and nitrogen follows:

As, hydrogen gas is present in excess. So, it is considered as an excess reagent.

Nitrogen gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of nitrogen gas is producing 2 moles of ammonia gas

So, 0.334 moles of nitrogen gas will produce = of ammonia gas.

Now, calculating the mass of ammonia gas by using equation 1, we get:

Moles of ammonia gas = 0.668 mol

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

Hence, the mass of ammonia produced in the reaction is 11.36 g

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2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

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