Halogens and Alkali react aggressively.
Answer:
2C3H8O + 9O2 ==> 6CO2 + 8H2O ... balanced equation
moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles
moles O2 = 31.8 g x 1 mol/31.9 g = 0.997 moles O2
Propanol is limiting based on the mol ratio in balance equation of 2 : 9
To find mass of O2 (excess reagent) left over, we will first find moles O2 used up.
moles O2 used = 0.0875 mol propanol x 9 mol O2/2 mol propanol = 0.394 moles O2 used
moles O2 left over = 0.997 mol - 0.394 mol = 0.603 mol O2 left
mass O2 left = 0.603 mol O2 x 32 g/mol = 19.3 g O2 left over
Answer:
Option 5. 179L
Explanation:
The reaction is:
2 HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂
The amount of collected CO₂ is 8 moles.
We apply the Ideal Gases Law at STP
STP are 1 atm of pressure and 273K of T°
P . V = n . R . T
1atm . V = 8 mol .0.082L.atm/mol.K . 273K
V = (8 mol .0.082L.atm/mol.K . 273K) / 1 atm = 179 L
Answer:
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