20 mol of NH, can be produce from 30 mol o H2
Answer:
A
Explanation:
The first option is the only option where new products are formed.
In the D option no new product is formed.
Same goes with option C.
In B no new products are formed.
Change is said to be taken place only with formation of new products from reactants.
Answer: When using 645 L /s of O2 in a temperature and pressure of 195°C, 0.88 atm respectively, we will get 0.355Kg /s NO
Explanation:
- First we review the equation that represents the oxidation process of the NH3 to NO.
4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)
- Second we gather the information what we are going to use in our calculations.
O2 Volume Rate = 645 L /s
Pressure = 0.88 atm
Temperature = 195°C + 273 = 468K
NO molecular weight = 30.01 g/mol
- Third, in order to calculate the amount of NO moles produced by 645L/ s of O2, we must find out, how many moles (n) are 645L O2 by using the general gas equation PV =n RT
Let´s keep in mind that using this equation our constant R is 0.08205Lxatm/Kxmol
PV =n RT
n= PV / RT
n= [ 0.88atm x 645L/s] / [ (0.08205 Lxatm/Kxmol) x 468K]
n= 14.781 moles /s of O2
-
Fourth, now by knowing the amount of moles of O2, we can use the equation to calculate how many moles of NO will be produced and then with the molecular weight, we will finally know the total mass per second .
14.781 moles /s of O2 x 4moles NO / 5 moles O2 x 30.01g NO / 1 mol NO x 1Kg NO /1000g NO = 0.355Kg /s NO
Answer:
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Explanation:
