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.
A. Potassium oxide
B. Calcium chloride
C. Magnesium nitride
D. Sodium hypochlorite
E. Potassium nitrate
Answer:
V₂ = 15.3
Explanation:
Given data:
Initial volume = 12.0 L
Initial temperature = 20°C
Final temperature =100°C
Final volume = ?
Solution:
First of all we will convert the temperature into kelvin.
20°C + 273 = 293 K
100°C + 273 = 373 K
Formula:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 12.0 L × 373 K / 293 k
V₂ = 4476 L.K /293 k
V₂ = 15.3
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
Get on mathpapa is shows you the answer and how to explain it
2H2O+O2--->2H2O2
8.5 gm H2O2=0.25 mole
hence H2O is also 0.25 mole i.e.4.5 gm
O2is 0.125 mole i.e.4 gm
