Problem 1: 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 calculate the amount of each reactant and compare it to the stoichiometric ratio in the balanced chemical equation.Given:- 15 grams of As- 25L of NaOHStep 1: Calculate the molar mass of As and NaOH.- Molar mass of As = 74.92 g/mol- Molar mass of NaOH = 39.99 g/mol + 15.999 g/mol + 1.008 g/mol = 56.996 g/molStep 2: Calculate the number of moles of As and NaOH.- Number of moles of As = 15 g / 74.92 g/mol = 0.201 mol- Number of moles of NaOH = 25L * 1 mol/L = 25 molStep 3: Determine the stoichiometric ratio between As and NaOH.- According to the balanced chemical equation, 2 moles of As react with 6 moles of NaOH.- Therefore, the stoichiometric ratio is 2:6 or 1:3.Step 4: Compare the number of moles of As and NaOH to the stoichiometric ratio.- The number of moles of As is 0.201 mol, which is less than the required amount of 0.201 mol * 3 = 0.603 mol of NaOH.- The number of moles of NaOH is 25 mol, which is more than the required amount of 0.201 mol * 2 = 0.402 mol of As.Therefore, the limiting reactant is arsenic (As).