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vova2212 [387]
4 years ago
5

A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to prod

uce carbon dioxide (CO2) and water (H2O). What mass of water is produced by the reaction of 7.72g of oxygen gas? Be sure your answer has the correct number of significant digits.
Chemistry
2 answers:
Olin [163]4 years ago
8 0

Answer:

The answer is 6.25g.

Explanation:

First create your balanced equation. This will give you the stoich ratios needed to answer the question:

2C8H18 + 25O2 → 16CO2 + 18H2O

Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:

7.72 g / 16 g/mol = 0.482 mol

Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:

x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2

x = 0.347 mol H2O

The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:

0.347 mol x 18 g/mol = 6.25g

vaieri [72.5K]4 years ago
4 0

Answer:

3.13 grams of water is produced by the reaction of 7.72 grams of oxygen gas.

Explanation:

2C_8H_{18} +25O_2\rightarrow 16CO_2+18H_2O

Moles of oxygen gas = \frac{7.72 g}{32 g/mol}=0.24125 mol

According to reaction , 25 moles of oxygen gas gives 18 moles of water.

Then 0.24125 moles of oxygen will give:

\frac{18}{25}\times 0.24125 mol=0.1737 mol

Mass of 0.1737 moles of water :

18 g/mol\times 0.1737 mol=3.1266 g\approx 3.13 g

3.13 grams of water is produced by the reaction of 7.72 grams of oxygen gas.

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. In a separate experiment, the molar mass of nicotine is found to be somewhere between 150 and 180 g/mol. Calculate the molar m
stealth61 [152]

Answer:

<h2>         162g/mol</h2>

Explanation:

The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:

<em>Nicotine has the formula   </em>C_xH_yN_z<em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>

  • <em>1.0 mol of CO₂</em>
  • <em>0.70 mol of H₂O</em>
  • <em>0.20 mol of NO₂</em>

<em>Assume that all the atoms in nicotine are present as products </em>

<h2>Solution</h2>

To find the empirical formula you need to find the moles of C, H, and N in each of the compound.

  • 1.0 mol of CO₂ has 1.0 mol of C
  • 0.70 mol of H₂O has 1.4 mol of H
  • 0.20 mol of NO₂ has 0.20 mol of N

Thus, the ratio of moles is:

  • C: 1.0
  • H: 1.4
  • N: 0.20

Divide all by the smallest number: 0.20

  • C: 1.0 / 0.20 = 5
  • H: 1.4 / 0.20 = 7
  • N: 0.20 / 0.20 = 1

Hence, the empirical formula is C₅H₇N

Find the mass of 1 mole of units of the empirical formula:

  • C:  5mol  × 12g/mol = 60g
  • H: 7mol × 1g/mol = 7 g
  • N: 1 mol × 14g/mol = 14g

Total mass = 60g + 7g + 14g = 81g

Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.

Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.

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Explanation:

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Under which circumstances are kp and kc equal for the reaction aa(g)+bb(g)⇌cc(g)+dd(g)?
slava [35]

Answer:

When the no. of moles of reactant gases (a + b) = the no. of moles of product gases (c + d).

Explanation:

  • The relation between Kp and Kc is:

Kp = Kc(RT)∧Δn.

  • So, Kp = Kc, when Δn = zero.
  • Δn is the difference between the no. of gas molecules in the products side and that in the reactants side.
  • So, Δn = zero, when the no. of molecules of gases in the products side = the no. of molecules of gases in the reactants side which means (a + b) = (c + d).
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