Question

what temperature, in degree Celcius could you collect 1.61 L of H2, at 679 mmHg, when reacting 3.47 g Mn and excess H2SO4. 2Mn+3H2SO4 = Mn2(SO4)3+3H2

Answer 1

Molar mass of Mn = 54.94 g/mol

mass of Mn = 3.47 g

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

= 3.47/54.94

= 0.0632 mol

From balanced chemical reaction, we see that

when 2 mol of Mn reacts, 3 mol of H2 is formed

mol of H2 formed = (3/2)* moles of Mn

= (3/2)*0.0632

= 0.0947 mol

we have:

P = 679.0 mm Hg

= (679.0/760) atm

= 0.8934 atm

V = 1.61 L

n = 0.0947 mol

we have below equation to be used:

P * V = n*R*T

0.8934 atm * 1.61 L = 0.0947 mol* 0.0821 atm.L/mol.K * T

T = 185 K

= (185-273) oC

= -88 oC

Answer: -88 oC