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2 years ago

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A mixture 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.

1 answer:

amid [387]2 years ago

4 0

Answer:

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%

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11 months ago

If 3.89 × 1024 atoms of a noble gas is collected and it has a mass of of 848 grams, this element is most likely a. He b. Ne c. A

bearhunter [10]

Answer:

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[α] = the specific rotation in degrees

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3 years ago

Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.

Shtirlitz [24]

Answer:

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Explanation:

To calculate de pH of an acid solution the formula is:

$pH = -Log ([H^{+}]) = 1$

were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.

(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.

(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:

$K_{w} =[H^{+} ][OH^{-}]=10^{-14}$

clearing the $[H^{+} ]$

$[H^{+} ]=\frac{10^{-14}}{[OH^{-}]}$

(d) is a weak base so it is necessary to solve the equilibrium first, knowing $Ka=3.24x10^{-8}$

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$Ka=3.24x10^{-8}=\frac{x^{2}}{5x10^{-8}-x}$

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${x^{2}={1.62x10^{-17}-3.24x10^{-8}x}$

$x=[H^{+}]=4.93x10^{-10}$

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3 years ago