2 answers:
Explanation:
The balance chemical equation is
2H2+O2—>2H2O
The limiting reactant is found by dividing the moles of each reactant by its coefficient
For H2=10.39/2=5.195
For O2=16.17/1= 16.17
Since H2 has the smaller value it’s the limiting reactant
The mass of the maximum amount of H2O is found by using the formula
moles= mass/molar mass
solving for mass gives
mass= moles*molar mass
(molar mass of H2O= 18g/mol)
mass= 5.195*18=93.51grams of H2O
Balanced chemical equation: 2H2 + O2 → 2H2O
The limiting reagent will be H2
All 10.39 mol of H2 will be used up produce 10.39 mol of H20
The excess reagent will be O2 and there will be 5.78 mols left over
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Answer:
I think it's B
hope this helps
litmus paper
Because it will just tell if the solution is acidic or basic, it won't tell the pH
Answer:

Explanation:
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 28.01 17.03
N₂ + 3H₂ ⟶ 2NH₃
m/g: 240.0
(a) Moles of NH₃

(b) Moles of N₂

(c) Mass of N₂
