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<span>You need to use some stoichiometry here. The only way to do that is if you're working in moles. Since you're given grams of Al, you can convert that moles by dividing by the molar mass.
Then from looking at the coefficients in your equation, you can see that for however many moles of Al react, the same numbers of moles of Fe will be produced, but only half as many moles of Al2O3 will be produced.
To go back to grams, multiply the moles of each product that you get by their molar masses!</span>
Answer:
N₃O₆
Step-by-step explanation:
Data:
EF = NO₂
MM = 138.02 g/mol
Calculations:
EF Mass = (14.01 + 32.00) u
EF Mass = 46.01 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ
n = MF Mass/EF Mass
n = 138.02 u/46.01 u
n = 3.000 ≈3
MF = (NO₂)₃
MF = N₃O₆
volume of H₂O = 7.2 L
Explanation:
The combustion reaction of methane (CH₄):
CH₄ + 2 O₂ → CO₂ + 2 H₂O
Now we calculate the number of moles of methane using the following formula:
number of moles = volume / 22.4 (L/mole)
number of moles of CH₄ = 3.6 / 22.4
number of moles of CH₄ = 0.16 moles
Taking in account the chemical reaction, we devise the following reasoning:
if 1 mole of CH₄ produce 2 moles of H₂O
then 0.16 moles of CH₄ produce X moles of H₂O
X = (0.16 × 2) / 1 = 0.32 moles of H₂O
And now we can calculate the volume of water (H₂O) produced by the reaction:
number of moles = volume / 22.4 (L/mole)
volume = number of moles × 22.4 (L/mole)
volume of H₂O = 0.32 × 22.4
volume of H₂O = 7.2 L
Learn more about:
combustion reaction
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