Calcium carbonate, CaCO3, in antacids reacts with HCl in the stomach to reduce acid reflux. 2HCl(aq) + CaCO3(s) → CO2(g) + H2O(l) + CaCl2(aq) How many liters of CO2 are produced at 752 mmHg and 24 °C from a 25.0-g sample of calcium carbonate?

Question

Calcium carbonate, [latex]CaCO_{3}[/latex], in antacids reacts with HCl in the stomach to reduce acid reflux.

[latex]2HCl(aq) + CaCO_{3}(s) → CO_{2}(g) + H_{2}O(l) + CaCl_{2}(aq)[/latex]

How many liters of [latex]CO_{2}[/latex] are produced at 752 mmHg and 24 °C from a 25.0-g sample of calcium carbonate?

Accepted Answer

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM	Given	Need	Connect
	25.0 g of [latex]CaCO_{3}[/latex] P = 752 mmHg T = 24 °C+ 273 = 297 K	V of [latex]CO_{2}(g)[/latex]	ideal gas law,PV= nRT, molar mass
	Equation
	[latex]2HCl(aq) + CaCO_{3}(s) → CO_{2}(g) + H_{2}O(l) + CaCl_{2}(aq)[/latex]

STEP 2 Write a plan to convert the given quantity to the needed moles.

[latex] grams of CaCO_{3} \boxed{Molar Mass} → moles of CaCO_{3} \boxed{Mole-mole factor } → moles of CO_{2}[/latex]

STEP 3 Write the equalities and conversion factors for molar mass and mole-mole factors.

[latex] \begin{array}{r c}\boxed{\begin{matrix} 1 mole of CaCO_{3} = 100.09 g of CaCO_{3} \\frac{100.09 g CaCO_{3}}{1 mole CaCO_{3}} ext{ and } \frac{1 mole CaCO_{3}}{100.09 g CaCO_{3}} \end{matrix}} & \boxed{\begin{matrix} 1 mole of CaCO_{3} = 1 mole of CO_{2} \ \frac{1 mole CaCO_{3}}{ 1 mole CO_{2}} ext{ and }\frac{1 mole CO_{2}}{1 mole CaCO_{3}} \end{matrix}}\end{array} [/latex]

STEP 4 Set up the problem to calculate moles of needed quantity.

[latex] 25.0 \cancel{g CaCO_{3}} imes \boxed{\frac{1 \cancel{mole CaCO_{3}}}{100.09 \cancel{ g CaCO_{3}}} } imes \boxed{\frac{1 mole CO_{2}}{1 \cancel{ mole CaCO_{3}}} } = 0.250 mole of CO_{2}[/latex]

STEP 5 Convert the moles of needed quantity to volume using the ideal gas law equation.

[latex] V=\frac{nRT}{P} [/latex]

[latex] V=\frac{0.250 \cancel{mole} imes \frac{62.4 L \cdot \cancel{mmHg}}{\cancel{mole} \cdot \cancel{K}} imes 297 \cancel{K}}{752 \cancel{mmHg}} = 6.16 L of CO_{2}[/latex]

Question 8.11: Calcium carbonate, CaCO3, in antacids reacts with HCl in the...

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