1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
Answer:
The 
for the reaction 
will be 4.69.
Explanation:
The given equation is A(B) = 2B(g)
to evaluate equilibrium constant for
![K_c=[B]^2[A]](https://tex.z-dn.net/?f=K_c%3D%5BB%5D%5E2%5BA%5D)
= 0.045
The reverse will be 
Then, ![K_c = \frac{[A]}{[B]^2}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5BA%5D%7D%7B%5BB%5D%5E2%7D)
= 
= 
The equilibrium constant for will be
= 4.69
Therefore, for the reaction will be 4.69.
Answer:
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