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:
54 g
Explanation:
1 mole of water = H2O
mass of 1 mole of H2O= mass of h2 + mass of o
= 2× mass of h +mass of o
= 2×1+16 =18 g
1 mole of water = 18g
3moles of water = 18×3g= 54g
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