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 answer of the questionis D Photons
I think it’s atoms because of google
Answer : The amount of heat released, 45.89 KJ
Solution :
Process involved in the calculation of heat released :
Now we have to calculate the amount of heat released.
![Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
Q = amount of heat released = ?
m = mass of water = 27 g
= specific heat of liquid water = 4.184 J/gk
= specific heat of solid water = 2.093 J/gk
= enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole
conversion : 
Now put all the given values in the above expression, we get
![Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]](https://tex.z-dn.net/?f=Q%3D%5B27g%5Ctimes%204.184J%2FgK%5Ctimes%20%28314-273%29k%5D%2B40700J%2B%5B27g%5Ctimes%202.093J%2FgK%5Ctimes%20%28273-263%29k%5D)
(1 KJ = 1000 J)
Therefore, the amount of heat released, 45.89 KJ
