Question

Sulfur trioxide gas can be removed from a gas stream, such as the exhaust gases of a coal-fired power plant, by reacting it with magnesium oxide. The reaction produces magnesium sulfate, according to the equation shown below. SO3 (g) + MgO (s) → MgSO4 (s) What volume of SO3 gas at 298 K and 0.832 atm could be captured by 1.50 × 102 kg of MgO?

Answer 1

Molar mass of MgO = 1*MM(Mg) + 1*MM(O)

= 1*24.31 + 1*16.0

= 40.31 g/mol

mass of MgO = 1.50*10^2 Kg = 150000 g

mol of MgO = (mass)/(molar mass)

= 150000/40.31

= 3721 mol

From balanced chemical reaction, we see that

when 1 mol of MgO reacts, 1 mol of SO3 is reacting

mol of SO3 recats = moles of MgO

= 3721 mol

we have:

P = 0.832 atm

n = 3721 mol

T = 298.0 K

we have below equation to be used:

P * V = n*R*T

0.832 atm * V = 3721 mol* 0.0821 atm.L/mol.K * 298 K

V = 1.09*10^5 L

Answer: 1.09*10^5 L