In ammonia, 14g of nitrogen combines completely with 3g of hydrogen. How many grams of hydrogen are needed to combined completely with 56 g of nitrogen?
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To solve this problem, we will use the concept of stoichiometry.
Step 1: Determine the molar masses of nitrogen (N) and hydrogen (H2)
- The molar mass of nitrogen (N) is 14.01 g/mol.
- The molar mass of hydrogen (H2) is 2.02 g/mol.
Step 2: Convert the given mass of nitrogen (56 g) into moles.
- Moles of nitrogen = Mass of nitrogen / Molar mass of nitrogen
= 56 g / 14.01 g/mol
= 4 moles of nitrogen
Step 3: Use the balanced chemical equation to determine the mole ratio between nitrogen and hydrogen.
- The balanced chemical equation for the formation of ammonia (NH3) is:
N2 + 3H2 -> 2NH3
- From the equation, we can see that for every 1 mole of nitrogen (N2) reacts, 3 moles of hydrogen (H2) are required.
Step 4: Calculate the moles of hydrogen required.
- Moles of hydrogen = Moles of nitrogen * Mole ratio
= 4 moles * 3 moles of H2 / 1 mole of N2
= 12 moles of H2
Step 5: Convert the moles of hydrogen into grams.
- Mass of hydrogen = Moles of hydrogen * Molar mass of hydrogen
= 12 moles * 2.02 g/mol
= 24.24 g
Therefore, 24.24 grams of hydrogen are needed to combine completely with 56 grams of nitrogen.