When calcium carbonate (CaCO_(3)) decomposes, it forms calcium oxide (CaO)and carbon dioxide (CO_(2)) gas. The-balanced chemical equation is: CaCO_(3)arrow CaO+CO_(2) How many grams of carbon dioxide (CO_(2)) will be produced from 75.0 grams of calcium carbonate (CaCO_(3)) 36.98 grams 38.98 grams 35.98 grams 32.98 grams

To determine the amount of carbon dioxide produced from the decomposition of calcium carbonate, we need to use stoichiometry based on the balanced chemical equation.<br /><br />The balanced chemical equation is:<br />\[ CaCO_{3} \rightarrow CaO + CO_{2} \]<br /><br />First, we need to calculate the molar masses of the compounds involved:<br />- Molar mass of \( CaCO_{3} \):<br /> - Calcium (Ca): 40.08 g/mol<br /> - Carbon (C): 12.01 g/mol<br /> - Oxygen (O): 16.00 g/mol (3 atoms of O)<br /><br /> \[ \text{Molar mass of } CaCO_{3} = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \, \text{g/mol} \]<br /><br />- Molar mass of \( CO_{2} \):<br /> - Carbon (C): 12.01 g/mol<br /> - Oxygen (O): 16.00 g/mol (2 atoms of O)<br /><br /> \[ \text{Molar mass of } CO_{2} = 12.01 + (2 \times 16.00) = 44.01 \, \text{g/mol} \]<br /><br />Next, we calculate the number of moles of \( CaCO_{3} \) in 75.0 grams:<br />\[ \text{Moles of } CaCO_{3} = \frac{75.0 \, \text{g}}{100.09 \, \text{g/mol}} = 0.749 \, \text{mol} \]<br /><br />According to the balanced equation, 1 mole of \( CaCO_{3} \) produces 1 mole of \( CO_{2} \). Therefore, 0.749 moles of \( CaCO_{3} \) will produce 0.749 moles of \( CO_{2} \).<br /><br />Finally, we convert the moles of \( CO_{2} \) to grams:<br />\[ \text{Mass of } CO_{2} = 0.749 \, \text{mol} \times 44.01 \, \text{g/mol} = 33.02 \, \text{g} \]<br /><br />Therefore, the amount of carbon dioxide produced from 75.0 grams of calcium carbonate is approximately 33.02 grams. However, since this value is not listed among the provided options, it seems there might be a mistake in the problem statement or the provided options. Based on the calculations, the correct answer should be approximately 33.02 grams.