Consider the chemical reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) to produce sodium chloride (NaCl) and water (H2O). Write a balanced chemical equation for this reaction and determine the mole ratio of NaOH and HCl required for complete reaction. If we start with 10 grams of NaOH and 20 grams of HCl, determine which is the limiting reagent, the amount of excess reagent left over, and the mass of NaCl and H2O produced.

Question

Consider the chemical reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) to produce sodium chloride (NaCl) and water (H2O). Write a balanced chemical equation for this reaction and determine the mole ratio of NaOH and HCl required for complete reaction. If we start with 10 grams of NaOH and 20 grams of HCl, determine which is the limiting reagent, the amount of exc

Answer 1

First, let's write the balanced chemical equation for this reaction:NaOH + HCl  NaCl + H2OThe mole ratio of NaOH to HCl is 1:1, as seen in the balanced equation.Now, let's determine the limiting reagent. First, we need to find the moles of each reactant:Moles of NaOH = mass / molar massMoles of NaOH = 10 g /  22.99 g/mol + 15.999 g/mol + 1.007 g/mol  = 10 g / 39.996 g/mol  0.250 molesMoles of HCl = mass / molar massMoles of HCl = 20 g /  1.007 g/mol + 35.453 g/mol  = 20 g / 36.460 g/mol  0.548 molesSince the mole ratio of NaOH to HCl is 1:1, NaOH is the limiting reagent because there are fewer moles of NaOH than HCl.Now, let's find the amount of excess HCl left over:Moles of excess HCl = moles of HCl - moles of NaOH = 0.548 moles - 0.250 moles = 0.298 molesMass of excess HCl = moles * molar mass = 0.298 moles * 36.460 g/mol  10.87 gNow, let's find the mass of NaCl and H2O produced:Moles of NaCl = moles of NaOH  since the mole ratio is 1:1  = 0.250 molesMass of NaCl = moles * molar mass = 0.250 moles *  22.99 g/mol + 35.453 g/mol  = 0.250 moles * 58.443 g/mol  14.61 gMoles of H2O = moles of NaOH  since the mole ratio is 1:1  = 0.250 molesMass of H2O = moles * molar mass = 0.250 moles *  2 * 1.007 g/mol + 15.999 g/mol  = 0.250 moles * 18.015 g/mol  4.50 gIn summary:- Limiting reagent: NaOH- Excess reagent: HCl, with 10.87 g left over- Mass of NaCl produced: 14.61 g- Mass of H2O produced: 4.50 g