Question

In the following reaction, what mass of mercury(II) oxide, HgO, would be required to produce 199...

In the following reaction, what mass of mercury(II) oxide, HgO, would be required to produce 199 L of oxygen, O2, measured at STP?
2HgO(s) --> 2Hg(s)+O2(g)

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Homework Answers

Answer #1

AT STP, 1 mol of any gas occupy 22.4 L

So,

moles of O2 produced = volume of O2 / molar volume

= 199 / 22.4

= 8.884 mol

From reaction,

moles of HgO = 2*moles of O2

= 2*8.884 mol

= 17.768 mol

Molar mass of HgO = 1*MM(Hg) + 1*MM(O)

= 1*200.6 + 1*16.0

= 216.6 g/mol

we have below equation to be used:

mass of HgO,

m = number of mol * molar mass

= 17.768 mol * 216.6 g/mol

= 3.85*10^3 g

Answer: 3.85*10^3 g

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