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Misha Larkins [42]
3 years ago
11

using the equation, c5h12 8o2 arrow 5co2 6h2o, if 108 g of water are produced, how many grams of oxygen were consumed?

Chemistry
1 answer:
vfiekz [6]3 years ago
8 0
Molar mass:

H₂O = 18.0 g/mol

O₂ = 32.0 g/mol

C₅H₁₂ + 8 O₂ -> 5 CO₂ + 6 H₂<span>O
</span>
8 x (32 g )<span> ------------ 6 x (18 g )</span>
mass O₂ ------------ 108 g H₂O

mass O₂ = 108 x 8 x 32 / 6 x 18

mass O₂ = 27648 / 108

mass O₂ =<span> 256 g</span>

<span>hope this helps!</span>
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3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

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Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

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4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

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2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

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b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

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