Problem 1: 2As+6NaOHarrow 2Na_(3)AsO_(3)+3H_(2) Given: 15 grams of As and 1 point 25L NaOH Find: the grams of H_(2) that can be produced? 0.60gH_(2) 1.12gH_(2) 45gH_(2) 25gH_(2) 10gH_(2)

To solve this problem, we need to determine the limiting reactant and then calculate the amount of hydrogen gas (H2) produced.<br /><br />Given information:<br />- 15 grams of arsenic (As)<br />- 25 liters of sodium hydroxide (NaOH)<br /><br />Step 1: Calculate the moles of As and NaOH.<br />Moles of As = 15 g / (74.92 g/mol) = 0.200 mol<br />Moles of NaOH = 25 L × (4.00 mol/L) = 100 mol<br /><br />Step 2: Determine the limiting reactant.<br />The balanced chemical equation is:<br />2As + 6NaOH → 2Na3AsO3 + 3H2<br /><br />From the balanced equation, we can see that 2 moles of As react with 6 moles of NaOH to produce 3 moles of H2.<br /><br />Moles of H2 produced from As:<br />0.200 mol As × (3 mol H2 / 2 mol As) = 0.300 mol H2<br /><br />Moles of H2 produced from NaOH:<br />100 mol NaOH × (3 mol H2 / 6 mol NaOH) = 50 mol H2<br /><br />Since the moles of H2 produced from As (0.300 mol) is less than the moles of H2 produced from NaOH (50 mol), As is the limiting reactant.<br /><br />Step 3: Calculate the grams of H2 produced.<br />Moles of H2 produced = 0.300 mol<br />Grams of H2 produced = 0.300 mol × (2.016 g/mol) = 0.6048 g<br /><br />Therefore, the grams of H2 that can be produced is 0.60 g.<br /><br />The correct answer is: $0.60gH_{2}$