Question

What mass of h2o is required to form 1.6 l of o2 at a temperature of 325 k and a pressure of 0.995 atm ?

Answer 1

2.14902 g

Step-by-step 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

Answer 2

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

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