Problem 2: 2As+6NaOHarrow 2Na_(3)AsO_(3)+3H_(2) Given: 15 grams of As and 25L NaOH. Which is the excess? arsenic sodium hydroxide sodium arsenite hydrogen gas

To determine which reactant is in excess, we need to calculate the amount of product formed from each reactant and compare them.<br /><br />First, let's calculate the amount of sodium arsenite (Na3AsO3) formed from arsenic (As):<br /><br />Given: 15 grams of As<br /><br />Molar mass of As = 74.92 g/mol<br /><br />Number of moles of As = 15 g / 74.92 g/mol = 0.200 mol<br /><br />According to the balanced chemical equation, 2 moles of As produce 2 moles of Na3AsO3. Therefore, 0.200 moles of As will produce 0.200 moles of Na3AsO3.<br /><br />Next, let's calculate the amount of sodium arsenite (Na3AsO3) formed from sodium hydroxide (NaOH):<br /><br />Given: 25 L of NaOH<br /><br />Molar volume of NaOH = 25 L<br /><br />Molar concentration of NaOH = 1 mol/L (assuming 1 M solution)<br /><br />Number of moles of NaOH = 25 L * 1 mol/L = 25 mol<br /><br />According to the balanced chemical equation, 6 moles of NaOH produce 2 moles of Na3AsO3. Therefore, 25 moles of NaOH will produce (25/6) * 2 = 8.33 moles of Na3AsO3.<br /><br />Comparing the amounts of Na3AsO3 formed from each reactant, we can see that the amount formed from arsenic (0.200 moles) is less than the amount formed from sodium hydroxide (8.33 moles). Therefore, arsenic is the limiting reactant, and sodium hydroxide is in excess.<br /><br />Answer: sodium hydroxide