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Jobisdone [24]
1 year ago
7

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

tion isCH4(g)+O2(g)→CO2(g)+H2O(g) This type of reaction is referred to as a complete combustion reaction.What mass of water is produced from the complete combustion of 5.90×10−3 g of methane?Express your answer with the appropriate units.
Chemistry
1 answer:
slega [8]1 year ago
5 0
<h2>Answer:</h2>1.33*10^{-2}grams

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}

Determine the mass of water produced

\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

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Synthesis gas consists of a mixture of H2 and CO. Calculate at 101kpa is compressed at constant temperature from 7.20 dm3 to 4.2
bazaltina [42]

Answer:

The final pressure is 173 kPa.

Explanation:

A mixture of gases at an initial pressure of  P₁ = 101 kPa is compressed from an initial volume V₁ = 7.20 dm³ to a final volume V₂ = 4.21 dm³. If we suppose an ideal gas behavior and that temperature remains constant, we can apply Boyle's law to find out the final pressure P₂:

P₁ . V₁ = P₂ . V₂

P_{2}=\frac{P_{1}.V_{1}}{V_{2}}=\frac{101kPa.7.20dm^{3} }{4.21dm^{3} }  =173kPa

5 0
4 years ago
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Which of these is an example of a way someone can help the environment? A. Increase the amount of travel done by car. B. Increas
HACTEHA [7]

Answer:  D. Increase the amount of travel done by bike.

Explanation:

Environment protection has become a concerned issue for the human society because of the fact that certain human activities are deteriorating the quality of air, water, and soil by liberating wastes into the environment.

Among the options given,  D. Increase the amount of travel done by bike. is the correct option. This is due to the fact that use of bike instead of using automobile like car will prevent the large amount of emission of harmful gases which are capable of polluting the air, water and soil can be restricted.  

4 0
4 years ago
What mass of iodine would react with 48 grams of magnesium to make magnesium iodide? Mg + I2 -&gt; MgI2
mrs_skeptik [129]
1) Chemical equation:

Mg + I2 ---> Mg I2

2) molar ratios

1 mol Mg : 1 mol I2 : 1 mol Mg I2

3) Calculate the number of moles of Mg in 48 grams

Atomic mass of Mg: 24.3 g/mol

number of moles = mass in grams / atomic mass = 48 g / 24.3 g/mol = 1.975 mol Mg

4) Use a proportion with the molar ratios

1 mol I2 / 1 mol Mg = x / 1.975 mol Mg =>

=> x = 1 mol I2 * 1.975 mol Mg / 1 mol Mg = 1.975 mol I2

5) Convert 1.975 mol I2 to grams

molar mass of I2 = 2 * 126.9 g/mol = 253.8 g/mol

mass = number of moles * molar mass = 1.975 mol * 253.8 g/mol = 501.255 g

Answer: 501 grams
6 0
3 years ago
Which statement about erosion is true? Most erosion occurs too slowly to observe. Most erosion occurs too quickly to observe, Er
olganol [36]

Answer:

Erosion can happen very quickly or very slowly.

e.g splash erosion occurs slowly while gulley erosion occurs very fast.

7 0
3 years ago
If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone
GREYUIT [131]

Answer:

m of NH3 = 6.46 g

Explanation:

First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:

NH₃ + O₂ ---------> NO + H₂O

Now, let's balance the equation:

4NH₃ + 5O₂ ---------> 4NO + 6H₂O

Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:

MM NH3 = 17 g/mol

MM O2 = 32 g/mol

moles NH3 = 18.1 / 17 = 1.06 moles

moles O2 = 27.2 / 32 = 0.85 moles

Now, let's compare these moles with the theorical moles that the balanced equation gave:

4 moles NH3 --------> 5 moles O2

1.06 moles ----------> X

X = 1.06 * 5 / 4 = 1.325 moles of O2

These means in order to  NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.

Now, let's see how many grams in excess we have left after the reaction is complete.

4 moles NH3 --------> 5 moles O2

X moles NH3 ----------> 0.85 moles

X = 0.85 * 4 / 5 = 0.68 moles of NH3

This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:

moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles

Finally the mass:

m = 0.38 * 17

<em>m = 6.46 g of NH3</em>

8 0
3 years ago
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