The equation for a reaction that can be used to manufacture lithium carbonate is 2Li_(2)O_(2)+2CO_(2)arrow 2Li_(2)CO_(3)+O_(2) What is the atom economy for this reaction? A 58.9 B 62.1 C 69.8 (D) 82.2

To calculate the atom economy for the given reaction, we need to follow these steps:<br /><br />1. Determine the molecular formula for each reactant and product involved in the reaction.<br />2. Calculate the molar mass of each reactant and product.<br />3. Calculate the total mass of all reactants and the total mass of all products.<br />4. Calculate the atom economy using the formula: Atom Economy = (Total mass of desired product / Total mass of all reactants) x 100%.<br /><br />Let's go through each step:<br /><br />1. Reactants: $2Li_{2}O_{2}$ and $2CO_{2}$<br /> - $Li_{2}O_{2}$: Lithium (Li) and Oxygen (O)<br /> - $CO_{2}$: Carbon (C) and Oxygen (O)<br /><br /> Products: $2Li_{2}CO_{3}$ and $O_{2}$<br /> - $Li_{2}CO_{3}$: Lithium (Li), Carbon (C), and Oxygen (O)<br /> - $O_{2}$: Oxygen (O)<br /><br />2. Molar masses:<br /> - $Li_{2}O_{2}$: 2*(2.7) + 2*(16) = 55.4 g/mol<br /> - $CO_{2}$: 12.01 + 2*(16) = 44.01 g/mol<br /> - $Li_{2}CO_{3}$: 2*(2.7) + 12.01 + 3*(16) = 73.89 g/mol<br /> - $O_{2}$: 2*(16) = 32.00 g/mol<br /><br />3. Total mass of all reactants:<br /> - $2Li_{2}O_{2}$: 2*(55.4) = 110.8 g/mol<br /> - $2CO_{2}$: 2*(44.01) = 88.02 g/mol<br /> - Total mass of all reactants = 110.8 + 88.02 = 198.82 g/mol<br /><br />4. Total mass of all products:<br /> - $2Li_{2}CO_{3}$: 2*(73.89) = 147.78 g/mol<br /> - $O_{2}$: 32.00 g/mol<br /> - Total mass of all products = 147.78 + 32.00 = 179.78 g/mol<br /><br />5. Atom economy:<br /> - Total mass of desired product = 147.78 g/mol<br /> - Total mass of all reactants = 198.82 g/mol<br /> - Atom economy = (147.78 / 198.82) x 100% = 74.1%<br /><br />Therefore, the correct answer is not listed among the options provided.