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Lubov Fominskaja [6]
3 years ago
14

Gaseous carbon monoxide reacts with hydrogen gas to form gaseous methane (ch4) and liquid water. express your answer as a chemic

al equation. identify all of the phases in your answer.
Chemistry
1 answer:
laiz [17]3 years ago
8 0
The formula for the compounds in the reaction are as follows with the respective states 
Carbon monoxide - CO (g)
hydrogen - H₂ (g)
methane - CH₄(g)
water - H₂O (l)
reaction of carbon monoxide with hydrogen gas gives rise to methane and water
the balanced chemical equation for the above reaction is as follows
CO(g) + 3H₂(g)  --> CH₄(g) + H₂O(l)
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Covalent Bond

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Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou
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Answer:

4.36~g~XY

Explanation:

In this case, we can start with the reaction:

2X + Y_2~->~2XY

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X

4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

\frac{0.04~mol~X}{1}=~0.04

\frac{0.0875~mol~Y_2}{2}=0.04375

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol Y_2 = 48 g Y_2 (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

I hope it helps!

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