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What volume of carbon dioxide will 3.26 g of antacid made of calcium carbonate produce at 37.0 °C and 1.00 atm in the stomach according to the following reaction?
CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Respuesta :

Answer:

0.830 mL of carbon dioxide will 3.26 g of antacid made of calcium carbonate produce at 37.0 °C and 1.00 atm.

Explanation:

[tex]CaCO_3 (s) + 2 HCl (aq)\rightarrow CaCl_2 (aq) + H_2O (l) + CO_2 (g) [/tex]

Moles of calcium carbonate = 3.26 g

Moles of calcium carbonate = [tex]\frac{3.26 g}{100 g/mol}=0.0326 mol[/tex]

According to reaction, 1 mol of calcium carbonate produces 1 mol of carbon dioxide gas, then 0.0326 moles of calcium carbonate will give :

[tex]\frac{1}{1}\times 0.0326 mol=0.0326 mol[/tex] of carbon dioxide.

Pressure of the gas = P = 1.00 atm

Temperature of the gas = T = 37.0°C = 37.0+ 273 K = 310 K

Volume of the gas = V

Moles of gas = n = 0.0326 mol

[tex]PV=nRT[/tex] ( ideal gas equation )

[tex]V=\frac{nRT}{P}=\frac{0.0326 mol\times 0.0821 atm J/mol K\times 310 K}{1.00 atm}=0.830 mL[/tex]

0.830 mL of carbon dioxide will 3.26 g of antacid made of calcium carbonate produce at 37.0 °C and 1.00 atm.

pstnonsonjoku

The volume occupied by the gas at 37.0 °C and 1.00 atm  is 0.82 L.

The equation of the reaction is;

CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + H₂O (l) + CO₂ (g)

Number of moles of carbonate =  3.26 g /100 g/mol = 0.0326 moles

Since 1 mole of carbonate yields 1 mole of carbon dioxide

Hence  0.0326 moles of carbonate yields  0.0326 moles of carbon dioxide

If 1 mole of carbon dioxide occupy 22.4 L at STP

0.0326 moles of carbon dioxide occupies 0.0326 moles ×  22.4 L/1 mole

= 0.73 L at STP

At 37.0 °C and 1.00 atm;

T1 = 273 K

P1 = 1 atm

V1 =  0.73 L

T2 = 37.0 °C + 273 = 310 K

P2 = 1 atm

V2 = ?

P1V1/T1 =P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 =  1 atm × 0.73 L × 310 K/1 atm × 273 K

V2 = 0.82 L

Learn more: https://brainly.com/question/9352088

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