Answer:
There can be prepared 693.512 kg of acrylonitrile
Explanation:
Propene has molar mass of 42.08g/mole
ammonia has molar mass of 17.03g/ mole
O2 has molar mass of 2*16 = 32g/mole
acrylonitrile has molar mass of 53.06g /mole
<u>Step 1</u>: finding limiting reagent
2C3H6 + 2NH3 + 3O2 → 2C3H3N + 6H2O
We see that the ratio 2:2:3:2:6 is
This means that for 2 mole C3H6 we have 2 mole 2NH3 and 3 mole 3O2, also there will be produced 2 mole C3H3N and 6 mole H2O
<u>Step2</u> : Calculating moles
moles propene: 550000g / 42.08g/mole = 13070.34 moles
moles ammonia: 650000g/ 17.03g/mole = 38167.94 moles
moles O2: 900000g/ 32g/mole = 28125moles
The limiting reagent is propene:
⇒propene will be consumed completely
⇒ammonia will consume 13070.34 moles ⇒ there will remain 25097.6 moles
⇒oxygen will consume (2/3 * 13070.34) = 18750moles ⇒ 9375 moles will remain
<u>Step 3</u>: Calculating moles of acrylonitrile
There will be formed 13070.34 moles of acrylonitrile
mass acrylonitrile = 13070.34 moles * 53.06g/mole =693512.24g = 693.512kg
There can be prepared 693.512 kg of acrylonitrile
