Question

Use the References to access important values if needed for this question. What volume of oxygen gas is produced when 57.4 g of mercury(II) oxide reacts completely according to the following reaction at 25 °C and 1 atm? mercury(II) oxide (s) —>mercury (1) + oxygen(g) liters oxygen gas

Answer 1

1)
Balanced chemical equation is:
2 HgO(s) -> 2 Hg(l) + O2(g)

Molar mass of HgO,
MM = 1*MM(Hg) + 1*MM(O)
= 1*200.6 + 1*16.0
= 216.6 g/mol

mass of HgO = 57.4 g
mol of HgO = (mass)/(molar mass)
= 57.4/2.166*10^2
= 0.265 mol

According to balanced equation
mol of O2 formed = (1/2)* moles of HgO
= (1/2)*0.265
= 0.1325 mol

Given:
P = 1.0 atm
n = 0.1325 mol
T = 25.0 oC
= (25.0+273) K
= 298 K

use:
P * V = n*R*T
1 atm * V = 0.1325 mol* 0.08206 atm.L/mol.K * 298 K
V = 3.2401 L
Answer: 3.24 L

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