To determine the mass of dihydrogen monoxide produced when 40.2 g of oxygen reacts with 10.7 g of hydrogen to produce dihydrogen monoxide, we need to follow the stoichiometry of the reaction.
The balanced chemical equation for the reaction is:
2H₂ + O₂ → 2H₂O
From the equation, we can see that 1 mole of O₂ reacts with 2 moles of H₂ to produce 2 moles of H₂O.
First, we need to find the limiting reactant to determine the amount of dihydrogen monoxide produced. The molar mass of O₂ is 32 g/mol, and the molar mass of H₂ is 2 g/mol56.
Number of moles of hydrogen (H₂):
10.7 g / 2 g/mol = 5.35 mol
Determine the limiting reactant:
Since 1.26 mol of O₂ is needed to react with 2.52 mol of H₂, and we have 1.26 mol of O₂ available, O₂ is the limiting reactant.
Calculate the mass of dihydrogen monoxide produced:
From the balanced equation, 2 moles of H₂O are produced for every 1 mole of O₂ consumed.
1.26 mol of O₂ will produce 1.26 mol of H₂O.
1.26 mol x (2 mol H₂O / 1 mol O₂) = 2.52 mol H₂O
Now, calculate the mass of dihydrogen monoxide produced:
Molar mass of H₂O = 18 g/mol
Mass of H₂O produced = 2.52 mol x 18 g/mol = 45.36 g
Therefore, 45.36 grams of dihydrogen monoxide would be produced in this reaction.
To determine how much of the excess reactant remains after the reaction, we first need to identify the limiting reactant, which in this case is oxygen (O₂) as calculated previously456.
Given that oxygen is the limiting reactant, we can find out how much of the excess reactant, which is hydrogen (H₂), remains unused by subtracting the amount of hydrogen that reacted from the initial amount245.
Calculating the moles of hydrogen that reacted:
1.26 mol of O₂ reacts with 2.52 mol of H₂ based on the stoichiometry of the reaction.
1.26 mol of O₂ x (2 mol H₂ / 1 mol O₂) = 2.52 mol H₂ reacted
Next, we subtract the moles of hydrogen that reacted from the initial moles of hydrogen to find out how much of the excess reactant remains:
Initial moles of hydrogen = 5.35 mol
Moles of hydrogen reacted = 2.52 mol
Remaining moles of excess hydrogen = 5.35 mol - 2.52 mol = 2.83 mol
Finally, we can convert the remaining moles of excess hydrogen to grams if needed by multiplying by the molar mass of hydrogen (2 g/mol):
2.83 mol x 2 g/mol = 5.66 grams
Therefore, after the reaction, 5.66 grams of excess hydrogen would remain unused.
To determine the mass of ammonia produced when 40.2 g of nitrogen reacts with 10.7 g of hydrogen to produce ammonia (NH₃), we need to follow the stoichiometry of the reaction.
The balanced chemical equation for the reaction is:
N₂ + 3H₂ → 2NH₃
From the equation, we can see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
First, we need to find the limiting reactant to determine the amount of ammonia produced6. The molar mass of N₂ is 28 g/mol, and the molar mass of H₂ is 2 g/mol6.
Number of moles of hydrogen (H₂):
10.7 g / 2 g/mol = 5.35 mol
Determine the limiting reactant:
Since 1.4357 mol of N₂ is needed to react with 4.3071 mol of H₂, and we have 1.4357 mol of N₂ available, N₂ is the limiting reactant.
Calculate the mass of ammonia produced:
From the balanced equation, 2 moles of NH₃ are produced for every 1 mole of N₂ consumed.
1.4357 mol of N₂ will produce 2.8714 mol of NH₃.
1.4357 mol x (2 mol NH₃ / 1 mol N₂) = 2.8714 mol NH₃
Now, calculate the mass of ammonia produced:
Molar mass of NH₃ = 17 g/mol
Mass of NH₃ produced = 2.8714 mol x 17 g/mol = 48.8 g
Therefore, 48.8 grams of ammonia would be produced in this reaction.