Amixture of calcium carbonate, caco3, and barium carbonate, baco3, weighing 5.40 g reacts fully with hydrochloric acid, hcl(aq), to generate 1.39 l co2(g), mea sured at 50°c and 0.904 atm pressure. calculate the percentages by mass of caco3 and baco3 in the original mixture.

CaCO₃ = 85.18%

BaCO₃ = 14.82%

Explanation:

The acid will react with the salts, the partial reactions are:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O

So, the total amount of CaCO₃ BaCO₃ will form the CO₂.

Using the ideal gas law to calculate the number of moles of CO₂:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature (50ºC + 273 = 323 K).

0.904*1.39 = n*0.082*323

26.486n = 1.25656

n = 0.05 mol

So, the number of moles of the mixture is 0.05 mol.

The molar masses of the components are:

CaCO₃ = 40 g/mol of Ca + 12 g/mol of C + 3*16 g/mol of O = 100 g/mol

BaCO₃ = 137.3 g/mol of Ba + 12 g/mol of C + 3*16 g/mol of O = 197.3 g/mol

Let's call x the number of moles of CaCO₃ and y the number of moles of BaCO₃, so:

100x + 197.3y = 5.4

x + y = 0.05 mol

y = 0.05 - x

100x + 197.3*(0.05 - x) = 5.4

100x - 197.3x = 5.4 - 9.865

97.3x = 4.465

x = 0.046 mol of CaCO₃

y = 0.004 mol of BaCO₃

So, the masses are:

CaCO₃ = 100* 0.046 = 4.60 g

BaCO₃ = 137.3*0.004 = 0.80 g

The percentages in the mixture are:

CaCO₃ = (4.60/5.40)*100% = 85.18%

BaCO₃ = (0.80/5.40)*100% = 14.82%

Answer is: -1300 kj/mol.p₄(s) + 3o₂(g) → p₄o₆(s), ∆h(p₄o₆(s))  = -1640,1 kj/mol.p₄(s) + 5o₂(g) → p₄o₁₀(s), ∆h(p₄o₁₀(s))  = -2940,1 kj/mol.δh(o₂) = 0 kj/mol.δh(reaction) = δh(reactants) - δh(product of reaction)δh(reaction) =  δh(p₄o₁₀) -    δh(p₄o₆).δh(reaction) = -2940,1 kj/mol  - (-1640,1 kj/mol) = -1300 kj/mol.