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3) The volume of ammonia produced when 12g of Hydrogen combine with excess nitrogen.

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Answer:

The balanced chemical reaction to form ammonia is-

N2 +3H2->2NH3

From the above reaction we can say 3 moles of hydrogen gas(h2) are required to produce 2 moles of ammonia (NH3).

To find the amount of ammonia produced when 12g of hydrogen reacts with excess nitrogen, we need to first convert the mass of hydrogen to moles using its molar mass. The molar mass of hydrogen is 1.008 g/mol, so:

12 g H2 × (1 mol H2 / 1.008 g H2) = 11.905 mol H2

Since the nitrogen is in excess, all 11.905 moles of hydrogen will react. This will produce:

2 mol NH3 / 3 mol H2 × 11.905 mol H2 = 7.937 mol NH3

Finally, we can convert the moles of ammonia to volume using the ideal gas law. Assuming standard temperature and pressure (STP) of 0°C and 1 atm, respectively, the molar volume of an ideal gas is 22.4 L/mol. Therefore:

7.937 mol NH3 × 22.4 L/mol = 177.8 L NH3

So, the volume of ammonia produced when 12g of hydrogen combines with excess nitrogen is 177.8 L.

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