Respuesta :
V: 1,6L = 1,6 dm³
T: 325 K
p: 0,995 atm = (0,995×1013)hPa = 1007,935 hPa
R = 83,14 hPa×dm³/mol×K
n: ?
..................
pV = nRT
n = pV/RT
n = (1007,935×1,6)/(83,14×325)
n = 1612,696/27020,5
n = 0,0597 mol O₂
________________
2H₂O -------> O₂ + 2H₂
36g..............1 mol.........
36g H₂O ------ 1 mol O₂
Xg --------------- 0,0597 mol O₂
X = 36×0,0597 = 2,1492g of water is required
:•)
T: 325 K
p: 0,995 atm = (0,995×1013)hPa = 1007,935 hPa
R = 83,14 hPa×dm³/mol×K
n: ?
..................
pV = nRT
n = pV/RT
n = (1007,935×1,6)/(83,14×325)
n = 1612,696/27020,5
n = 0,0597 mol O₂
________________
2H₂O -------> O₂ + 2H₂
36g..............1 mol.........
36g H₂O ------ 1 mol O₂
Xg --------------- 0,0597 mol O₂
X = 36×0,0597 = 2,1492g of water is required
:•)
Answer:
2.14902 g
Explanation:
First thing's first, we have to write out the balanced chemical equation. This is given as;
2H2O --> O2 + 2H2
From the equation above, we can tell that 2 moles of H2O forms 1 mole of O2.
Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."
This means; 3.2L ( 2 * 1.6) OF H2O would form 1.6L of O2.
Other parameters given are;
Pressure = 0.995atm
Temperature = 325K
Mass = ?
But mass = number of moles * Molar mass
Molar mass of H2O = 18g
Number of moles?
From ideal gas equation;
PV = nRT
n = PV / RT
R = gas constant = 0.082057 L atm mol-1 K-1
n = (0.995 * 3.2) / (0.082057 * 325)
n = 3.184 / 26.669
n = 0.11939 moles
Mass = 0.11939 * 18 = 2.14902 g
