Question

A 2.00 g sample of HgO decomposes according to the reaction below to yield oxygen gas, which is collected at 300 K and 0.910 atm. What volume of oxygen is collected?

2 HgO (s) → 2 Hg (l) + O2 (g)

Answer #1

Molar mass of HgO = At.mass of Hg + At.mass of O

= 200.6 + 16

= 216.6 g/mol

Given mass of HgO = 2.00 g

So number of moles of HgO, n = mass/molar mass

= 2.00 / 216.6

= 0.00923 mol

2 HgO (s) → 2 Hg (l) + O_{2} (g)

From the above balanced equation,

2 moles of HgO produces 1 mole of O_{2}

0.00923 moles of HgO produces M mole of O_{2}

M = (1x0.00923) / 2

= 0.00462 moles

Calculation of volume of O2 gas :

We know that PV = nRT

P = pressure = 0.910 atm

V = volume = ?

n = number of moles = 0.00462 mol

R = gas constant = 0.0821 Latm/(mol-K)

T = temperature = 300 K

Plug the values we get V = (nRT) / P

= (0.00462x0.0821x300) / 0.910

= 0.125 L

= 125 mL

Therefore the volume of oxygen collected is 125 mL

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