In the reaction below, 34.8 g of solid calcium carbonate is mixed with 51.6 g hydrochloric acid. How many grams of carbon dioxide will be produced?
CaCO3 + 2HCl -> CaCl2 + CO2+H2O
a. 12.3 g carbon dioxide
b. 30.6 g carbon dioxide
c. 7.7 g carbon dioxide
d. 31.1 g carbon dioxide
e. 15.3 g carbon dioxide
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To solve this problem, we need to determine the limiting reactant (the reactant that will be completely consumed) and then calculate the grams of carbon dioxide produced from that reactant.
First, let's calculate the number of moles of each reactant:
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 100.09 g/mol
Moles of CaCO3 = 34.8 g / 100.09 g/mol = 0.3477 mol
Molar mass of HCl = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
Moles of HCl = 51.6 g / 36.46 g/mol = 1.416 mol
Next, we need to determine the stoichiometric ratio between CaCO3 and CO2. From the balanced equation, we can see that for every 1 mole of CaCO3, 1 mole of CO2 is produced.
Since we have 0.3477 mol of CaCO3, the maximum amount of CO2 that can be produced is also 0.3477 mol.
Finally, we convert the moles of CO2 to grams using the molar mass of CO2:
Molar mass of CO2 = 12.01 g/mol (C) + 2(16.00 g/mol) (O) = 44.01 g/mol
Grams of CO2 = 0.3477 mol * 44.01 g/mol = 15.3 g
Therefore, the correct answer is (e) 15.3 g carbon dioxide.
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