Sulfur dioxide will react with water to form sulfurous acid (see balanced equation below).<\/p>\n\n\n\n
SO2(g) + H20() \u2192 H2SO3()<\/p>\n\n\n\n
What mass of sulfur dioxide is needed to prepare 27.86 g of H2SO3<\/p>\n\n\n\n
(b? Molar mass: SO2: 64.06 g\/mol<\/p>\n\n\n\n
Molar mass: H2SO3: 82.08 g\/mol<\/p>\n\n\n\n
a. 21.74 g b.0.4348 g c. 35.69 g d. 0.3394 g e. 27.86 g<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the mass of sulfur dioxide (SO\u2082) needed to prepare 27.86 g of sulfurous acid (H\u2082SO\u2083), we need to follow the steps of stoichiometric calculations. The chemical equation provided is:<\/p>\n\n\n\n [ First, calculate the number of moles of H\u2082SO\u2083 that corresponds to 27.86 grams. We use the molar mass of H\u2082SO\u2083, which is given as 82.08 g\/mol.<\/p>\n\n\n\n [ From the balanced chemical equation, we can see that the mole ratio between SO\u2082 and H\u2082SO\u2083 is 1:1. This means that for every 1 mole of H\u2082SO\u2083 produced, 1 mole of SO\u2082 is consumed.<\/p>\n\n\n\n So, the moles of SO\u2082 needed will be the same as the moles of H\u2082SO\u2083:<\/p>\n\n\n\n [ Next, we convert the moles of SO\u2082 to grams. Using the molar mass of SO\u2082, which is 64.06 g\/mol, we can find the mass of SO\u2082 required:<\/p>\n\n\n\n [ The correct answer is 21.74 g<\/strong>, which corresponds to option (a). Therefore, 21.74 grams of sulfur dioxide is needed to prepare 27.86 grams of sulfurous acid.<\/p>\n\n\n\n To summarize, we used the molar mass of H\u2082SO\u2083 to find the number of moles, then used stoichiometry to relate the moles of H\u2082SO\u2083 to the moles of SO\u2082, and finally converted moles of SO\u2082 to grams using its molar mass. This approach gives us the correct mass of SO\u2082 required to prepare the desired amount of H\u2082SO\u2083.<\/p>\n","protected":false},"excerpt":{"rendered":" Sulfur dioxide will react with water to form sulfurous acid (see balanced equation below). SO2(g) + H20() \u2192 H2SO3() What mass of sulfur dioxide is needed to prepare 27.86 g of H2SO3 (b? Molar mass: SO2: 64.06 g\/mol Molar mass: H2SO3: 82.08 g\/mol a. 21.74 g b.0.4348 g c. 35.69 g d. 0.3394 g e. 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\\text{SO}_2(g) + \\text{H}_2\\text{O}(l) \\rightarrow \\text{H}_2\\text{SO}_3(aq)
]<\/p>\n\n\n\nStep 1: Calculate moles of H\u2082SO\u2083<\/h3>\n\n\n\n
\\text{moles of H}_2\\text{SO}_3 = \\frac{\\text{mass of H}_2\\text{SO}_3}{\\text{molar mass of H}_2\\text{SO}_3} = \\frac{27.86 \\, \\text{g}}{82.08 \\, \\text{g\/mol}} = 0.3394 \\, \\text{mol}
]<\/p>\n\n\n\nStep 2: Use stoichiometry to find moles of SO\u2082<\/h3>\n\n\n\n
\\text{moles of SO}_2 = 0.3394 \\, \\text{mol}
]<\/p>\n\n\n\nStep 3: Calculate the mass of SO\u2082<\/h3>\n\n\n\n
\\text{mass of SO}_2 = \\text{moles of SO}_2 \\times \\text{molar mass of SO}_2 = 0.3394 \\, \\text{mol} \\times 64.06 \\, \\text{g\/mol} = 21.74 \\, \\text{g}
]<\/p>\n\n\n\nStep 4: Conclusion<\/h3>\n\n\n\n
Summary<\/h3>\n\n\n\n