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9966 [12]
3 years ago
6

If 2.32 g of ethanol reacts with 10.6 g of oxygen, how many moles of water are produced?

Chemistry
1 answer:
MAXImum [283]3 years ago
5 0
<span>0.151 moles Ethanol is C2H6O, and the balanced reaction is C2H6O + 3O2 ==> 2CO2 + 3H2O Now calculate the molar mass of C2H6O. Start with the atomic weights. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass C2H6O = 2 * 12.0107 + 6 * 1.00794 + 15.999 = 46.06804 g/mol Molar mass O2 = 2 * 15.999 = 31.998 g/mol Determine how many moles of each reactant is available. Moles C2H6O = 2.32 g / 46.06804 g/mol = 0.050360293 mol Moles O2 = 10.6 g / 31.998 g/mol = 0.331270704 mol Looking at the balanced equation and the number of moles of each reactant, it's obvious that the limiting reactant is C2H6O, and since each mole of C2H6O will produce 3 moles of H2O, then we just need to multiply the number of moles of C2H6O we have by 3, giving 0.050360293 mol * 3 = 0.151080879 moles And of course, round to 3 significant figures, giving 0.151 moles.</span>
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Sedbober [7]
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3 years ago
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anyanavicka [17]

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2 years ago
Select the lewis structure for xeo2f2 which correctly minimizes formal charges.
IrinaVladis [17]

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explanation and image attached

Explanation:

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Image Credit: UCLA

7 0
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tamaranim1 [39]

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8 0
2 years ago
Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0
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<span> Molar mass (H2)=2*1.0=2.0 g/mol
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From reaction        1 mol      1 mol
From problem      2.50 mol   1 .00mol

We can see that  excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.

</span>                                 H2(g) + F2(g) → 2 HF(g)
From reaction                      1 mol       2  mol
From problem                      1.00 mol  2.00mol

2.00 mol HF can be formed.

2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
4 0
3 years ago
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