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quester [9]
3 years ago
11

Missions to mars are feasible only once every two years ?

Chemistry
1 answer:
krek1111 [17]3 years ago
6 0

Answer:

Yes.

Explanation:

If you want to use the least fuel, then missions to Mars are feasible every two years.

You want to launch when Earth and Mars are close together, and this happens <em>about every 26 months</em>.

However, Earth travels faster than Mars, so the spaceship must launch when Mars is 44° <em>ahead </em>of Earth in its orbit.

The ship will then use the most efficient trajectory to arrive at Mars.

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Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc
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<u>Answer:</u> The percent yield of the water is 31.98 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

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Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol

The chemical equation for the combustion of methane is:

CH_4+2O_2\rightarrow CO_2+2H_2O

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = \frac{1}{2}\times 0.45=0.225mol of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = \frac{2}{2}\times 0.45=0.45 moles of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g

  • To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%

Hence, the percent yield of the water is 31.98 %

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