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Inessa05 [86]
3 years ago
9

During combustion, methane yields carbon dioxide and water. The unbalanced equation for this reaction is:CH4(g)+O2(g) → CO2(g)+

H2O(l)
What will the mole ratios for the balanced equation be? What coefficients are needed in order to balance the equation?
Chemistry
1 answer:
almond37 [142]3 years ago
4 0

Answer:

CH₄(g)  +  2O₂(g) → CO₂(g)  +  2H₂O(l)

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2

Explanation:

CH4(g)+O2(g) → CO2(g)+ H2O(l)

To balance a chemical equation, you must have the same mole of each element in both sides of the reaction (reactant side and product side)

CH₄(g)  +  2O₂(g) → CO₂(g)  +  2H₂O(l)

2 C

8 H

8 O

In  both side.

It takes 1 mole of methane to react with 2 mole of oxygen in order to produce 1 mol of dioxide and 2 mole of water.

Mole ratios for the balanced equation be:

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2  - We should compare each compound

1 mol methane → 2 mole of oxygen

2 mole of oxygen → 1 mol of methane

2 mole of oxygen → 1 mol of dioxide

1 mole of dioxide → 1 mol of Methane

2 mole of water → 2 mole of oxygen (the same as opposite)

1 mol methane →  2 mole of water

2 mol of water → 1 mol of methane

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Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

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7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

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1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

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= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

9) Since Fe2O3 is the limiting reactant, the ideal yield of Fe for the reaction is 61 g. If the actual reaction only gave us 25 g Fe. then the percent yield of Fe is

%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

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The air in a bicycle tire is bubbled through water and collected at 25 ∘C. If the total volume of gas collected is 5.30 L at a t
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<u>Answer:</u> The number of moles of gas in bicycle tire is 0.210 moles

<u>Explanation:</u>

To calculate the moles of gas, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 737 torr

V = Volume of the gas = 5.30 L

T = Temperature of the gas = 25^oC=[25+273]K=298K

R = Gas constant = 62.364\text{ L. Torr }mol^{-1}K^{-1}

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Putting values in above equation, we get:

737Torr\times 5.30L=n\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 298K\\\\n=\frac{737\times 5.30}{62.364\times 298}=0.210mol

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