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pogonyaev
2 years ago
6

What mass of water is produced from the complete combustion of 4.60×10-3 g of methane?

Chemistry
1 answer:
vladimir1956 [14]2 years ago
7 0

Answer:

=  1.03 x 10⁻² g H₂O

Explanation:

To find the mass of water, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles H₂O (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles H₂O to grams H₂O (via molar mass). The final answer should have 3 sig figs to reflect the given value (4.60 x 10⁻³ g).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Combustion of Methane:

1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

4.60 x 10⁻³ g CH₄          1 mole             2 moles H₂O          18.014 g
--------------------------  x  -----------------  x  ---------------------  x  ----------------  =
                                     16.043 g            1 mole CH₄             1 mole

=  0.0103 g H₂O

=  1.03 x 10⁻² g H₂O

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Combustion of hydrocarbons such as butane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosp
Dahasolnce [82]

Answer:

1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. V = 596L

Explanation:

Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:

C₄H₁₀ + O₂ → CO₂ + H₂O

1. The balanced chemical equation is:

C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. 0,360kg of butane are:

360g×\frac{1mol}{58,12g}=<em>6,19moles of butane</em>

These moles of butane are:

6,19moles of butane×\frac{4CO_2}{1molButane}= <em>24,8 moles CO₂</em>

Using V=nRT/P

Where:

n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).

Volume (V) is:

<em>V = 596L</em>

I hope it helps!

8 0
3 years ago
Given 40 grams of magnesium. How many mole is this
Dahasolnce [82]

To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.

We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'

\frac{40 g Mg}{1} x\frac{1 mole Mg}{24.305 g Mg}

In this, we can see that the unit 'grams' will cancel out to leave us with moles.

In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.

3 0
3 years ago
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Why are halogens so reactive?
TEA [102]
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3 0
3 years ago
Read 2 more answers
The limiting reactant, O2, can form up to 2.7 mol Al2O3. What mass of Al2O3 forms?
Margaret [11]

Answer:

280 g Al₂O₃

Explanation:

To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).

Molar Mass (Al₂O₃): 101.96 g/mol

2.7 moles Al₂O₃          101.96 g
------------------------  x  -------------------  = 275 g Al₂O₃  = 280 g Al₂O₃
                                     1 mole

5 0
2 years ago
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