Iodine is prepared both in the laboratory and commercially by adding Cl, (g) to an aqueous solution containing sodium iodide. 2 Nal(aq) + Cl\u00e2\u201a\u201a(g) \u00e2\u2020\u2019 1\u00e2\u201a\u201a (s) + 2 NaCl(aq) How many grams of sodium iodide, Nal, must be used to produce 60.0 g of iodine, I\u00e2\u201a\u201a? mass:<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the grams of sodium iodide (NaI) needed to produce 60.0 g of iodine (I\u2082), we’ll use stoichiometry, which involves balancing the chemical equation and converting between grams, moles, and other quantities. Here\u2019s the breakdown:<\/p>\n\n\n\n Step 1: Write the balanced chemical equation.<\/strong> [ 2 \\, \\text{NaI (aq)} + \\text{Cl}_2 \\, \\text{(g)} \\rightarrow \\text{I}_2 \\, \\text{(s)} + 2 \\, \\text{NaCl (aq)} ]<\/p>\n\n\n\n Step 2: Molar mass of iodine (I\u2082).<\/strong> Step 3: Use the mass of iodine to find moles of iodine.<\/strong> [ Step 4: Use the stoichiometry of the reaction.<\/strong> [ Step 5: Convert moles of NaI to grams.<\/strong> Now, we can calculate the mass of NaI required:<\/p>\n\n\n\n [ Final Answer:<\/strong> Iodine is prepared both in the laboratory and commercially by adding Cl, (g) to an aqueous solution containing sodium iodide. 2 Nal(aq) + Cl\u00e2\u201a\u201a(g) \u00e2\u2020\u2019 1\u00e2\u201a\u201a (s) + 2 NaCl(aq) How many grams of sodium iodide, Nal, must be used to produce 60.0 g of iodine, I\u00e2\u201a\u201a? mass: The Correct Answer and Explanation is : 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The given reaction is:<\/p>\n\n\n\n
The molar mass of iodine (I\u2082) is calculated by adding the molar masses of two iodine atoms:
[ 1 \\, \\text{I} = 126.9 \\, \\text{g\/mol} ]
So:
[ \\text{Molar mass of I}_2 = 2 \\times 126.9 \\, \\text{g\/mol} = 253.8 \\, \\text{g\/mol} ]<\/p>\n\n\n\n
We know the mass of iodine is 60.0 g. To find the number of moles of iodine, we divide the mass by the molar mass of iodine:<\/p>\n\n\n\n
\\text{Moles of I}_2 = \\frac{60.0 \\, \\text{g}}{253.8 \\, \\text{g\/mol}} = 0.236 \\, \\text{mol of I}_2
]<\/p>\n\n\n\n
From the balanced equation, 2 moles of NaI produce 1 mole of I\u2082. This means that for every mole of iodine produced, 2 moles of sodium iodide are required. Therefore, the moles of NaI required to produce 0.236 moles of iodine are:<\/p>\n\n\n\n
\\text{Moles of NaI} = 0.236 \\, \\text{mol I}_2 \\times \\frac{2 \\, \\text{mol NaI}}{1 \\, \\text{mol I}_2} = 0.472 \\, \\text{mol NaI}
]<\/p>\n\n\n\n
The molar mass of NaI is calculated by adding the molar mass of sodium (Na) and iodine (I):
[ \\text{Molar mass of NaI} = 22.99 \\, \\text{g\/mol} + 126.9 \\, \\text{g\/mol} = 149.89 \\, \\text{g\/mol} ]<\/p>\n\n\n\n
\\text{Mass of NaI} = 0.472 \\, \\text{mol NaI} \\times 149.89 \\, \\text{g\/mol} = 70.8 \\, \\text{g NaI}
]<\/p>\n\n\n\n
To produce 60.0 g of iodine (I\u2082), you need 70.8 grams of sodium iodide (NaI)<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"