39.25 g of water (H₂O)
Explanation:
We have the following chemical reaction:
2 H₂ + O₂ → 2 H₂O
Now we calculate the number of moles of each reactant:
number of moles = mass / molar weight
number of moles of H₂ = 14.8 / 2 = 7.4 moles
number of moles of O₂ = 34.8 / 32 = 1.09 moles
We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:
if 1 moles of O₂ produces 2 moles of H₂O
then 1.09 moles of O₂ produces X moles of H₂O
X = (1.09 × 2) / 1 = 2.18 moles of H₂O
mass = number of moles × molar weight
mass of H₂O = 2.18 × 18 = 39.25 g
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limiting reactant
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Answer:
The answer to your question is: V = 6.93 L
Explanation:
Data
N₂ = 5.6 g
Volume of NH₃ = ?
14 g of N ---------------- 1 mol
5.6 g ----------------------- x
x = (5.6 x 1) / 14 = 0.4 mol of N
Reaction
N₂ + 3H₂ ⇒ 2NH₃
1 mol of N₂ ---------------- 2 moles of NH₃
0.4 mol of N₂ -------------- x
x = (0.4 x 2) / 1
x = 0.8 mol of NH₃
Formula
PV = nRT
P = 5200 torr = 6.84 atm
V = ?
n = 0.8
R = 0.082 atm L/ mol °K
T = 450°C = 723°K
Substitution
V = (0.8)(0.082)(723) / 6.84
V = 6.93 L
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Answer:
It's C, direct and peripheral.
Explanation:
Just took the test
