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How many grams of hydrogen are necessary to react completely with 50.0 g of nitrogen in the reaction of the formation of ammonia?

N₂ + 3H₂ → 2NH₃
O 10.8 g H₂
O 15.0 g H₂
O 20.1 g H₂
O 41.2 g H₂

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Final answer:

10.7 grams of hydrogen are necessary to react completely with 50.0 g of nitrogen in the formation of ammonia.

Step-by-step explanation:

According to the balanced chemical equation, 1 molecule of nitrogen reacts with 3 molecules of hydrogen to produce 2 molecules of ammonia. To determine the mass of hydrogen that reacts completely with 50.0 g of nitrogen, we need to use mole-to-mole ratios and molar masses. The molar mass of nitrogen is 28.0 g/mol, while the molar mass of hydrogen is 2.0 g/mol. First, we convert the mass of nitrogen to moles using its molar mass:

50.0 g N2 * (1 mol N2/ 28.0 g N2) = 1.79 mol N2

Next, we use the mole-to-mole ratio from the balanced equation to find the moles of hydrogen:

1.79 mol N2 * (3 mol H2/ 1 mol N2) = 5.36 mol H2

Finally, we convert the moles of hydrogen to grams using its molar mass:

5.36 mol H2 * (2.0 g H2/ 1 mol H2) = 10.7 g H2

Therefore, 10.7 grams of hydrogen are necessary to react completely with 50.0 g of nitrogen.

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