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Chemical reaction: 2PCl₃ + O₂ → 2POCl₃.
m(PCl₃) = 194 g.
n(PCl₃) = m(PCl₃) ÷ M(PCl₃).
n(PCl₃) = 194 g ÷ 137,33 g/mol.
n(PCl₃) = 1,412 mol.
From chemical reaction: n(PCl₃) : n(POCl₃) = 1 : 1.
n(POCl₃) = 1,412 mol.
V(POCl₃) = n(POCl₃) · Vm.
V(POCl₃) = 1,412 mol · 22,4 mol/dm³.
V(POCl₃) = 31,64 dm³ at normal pressure and temperature.
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Answer:

V = 2.28 L

Explanation:

What volume of pocl3 gas can be produced in theory when 194 g of pcl3 gas reacts with excess oxygen? at STP

1. Determine moles PCl3.

The reaction

2PCl3+O2--------->PClO3

mole =mass/molecular mass

194 grams/137.3334 grams/mole  =

0.102 moles PCl3

2. Get the moles of POCl3 produced from 0.102 moles PCl3.

from the reaction above

2 moles PCl3 produce 2 moles POCl3

PClO3=0.102mole

3. Find the  volume PClO3 using ideal-gas equation: PV = nRT

where:

P = pressure = 1 atm (standard pressure)

V = volume ?

n = moles PClO3 = 0.102 moles PClO3

R = gas constant = 0.08206 L*atm/mol*K

T = temperature = 273.15 K (standard temperature)

(1 atm)*V = (0.102 moles)*(0.08206 L*atm/mol*K)*(273.15 K)

V = 2.28 L

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