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2 years ago

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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 ac

cording to the following reaction?
CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + H₂O (l) + CO₂ (g)

1 answer:

andrezito [222]2 years ago

7 0

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:

$CaCO_3 (s) + 2 HCl (aq)\rightarrow CaCl_2 (aq) + H_2O (l) + CO_2 (g)
$

Moles of calcium carbonate = 3.26 g

Moles of calcium carbonate = $\frac{3.26 g}{100 g/mol}=0.0326 mol$

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

$\frac{1}{1}\times 0.0326 mol=0.0326 mol$ 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

$PV=nRT$ ( ideal gas equation )

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

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.

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<u>Explanation:</u>

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$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

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<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

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To formulate the empirical formula, we need to follow some steps:

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kirza4 [7]

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<u></u>

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<h3>Further explanation</h3>

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