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

To determine the limiting reactant, we need to compare the amount reactant available to the amount required by the balanced chemical equation.<br /><br />Given:<br />- 15 grams of As<br />- 25 L of NaOH<br /><br />First, let's calculate the number of moles of each reactant.<br /><br />1. Calculate the number of moles of As:<br /> - Molar mass of As = 74.92 g/mol<br /> - Moles of As = 15 g / 74.92 g/mol = 0.2006 mol<br /><br />2. Calculate the number of moles of NaOH:<br /> - Molar mass of NaOH = 39.99 g/mol + 15.999 g/mol + 1.008 g/mol = 56.996 g/mol<br /> - Density of NaOH solution = 1.00 g/mL (assuming 1 M solution)<br /> - Volume of NaOH solution = 25 L = 25000 mL = 25000 g<br /> - Moles of NaOH = 25000 g / 56.996 g/mol = 438.5 mol<br /><br />Now, let's compare the mole ratio of each reactant to the balanced chemical equation:<br />\[ 2As + 6NaOH \rightarrow 2Na_{3}AsO_{3} + 3H_{2} \]<br /><br />From the balanced equation, we see that 2 moles of As react with 6 moles of NaOH. Therefore, the mole ratio is:<br />\[ \frac{6 \text{ moles of NaOH}}{2 \text{ moles of As}} = 3 \]<br /><br />Now, let's calculate the actual ratio of moles of NaOH to moles of As:<br />\[ \frac{438.5 \text{ moles of NaOH}}{0.2006 \text{ moles of As}} = 2195.5 \]<br /><br />Since the actual ratio (2195.5) is much greater than the required ratio (3), it means that there is an excess of NaOH. Therefore, the limiting react arsenic (As).<br /><br />So, the correct answer is:<br />**arsenic**