Question

Laboratory Manual for Decomposing Bong Soda 3. Potassium chlorate is heated to give potassium chloride and o pen as Ira 1.115-g sample of potassium chlorate is decomposed, how many milliliters of oxygen gas is released at STP 2 KCIO36) 2 KCK) - 30%) 4. Potassium metal reacts with water to give potassium hydroxide and hydrogen gas. If 75.5 mL of hydrogen gas is produced at STP, what is the mass of potassium metal that reacted? 2 KO) + 2 H206 — 2 KOH(aq) + H (8) 5. (optional) The Solvay process is used to manufacture baking soda, NaHCO,. In the process, CO2, NH3, H2O, and NaCl react to produce baking soda. Ir 15.0 L CO2 and 10.0L NH3 react at STP, with excess water and sodium chloride, what is the limiting reactant? Calculate the mass of baking soda produced. CO2(g) + NH3(e) + H2O(l) + NaCl(s) - NaHCO3(s) + NH4Cl(aq) Limiting Reactant Mass NaHCO3

Answer 1

3) No of mol of KClO3 reacted = 1.115/122.55 = 0.0091 mol

from equation, 2 mol KClO3 = 3 mol O2

No of mol of O2 formed = 0.0091*3/2 = 0.01365 mol

volume of O2 formed(V) = nRT/P

    n = 0.01365 mol

    R = 0.0821 l.atm.k-1.mol-1

    T = 273.15 k   (at STP)

    P = 1 atm

V = 0.01365*0.0821*273.15/1

   = 0.306 L

   = 306 ml

4) from equation, 2 mol K = 1 mol H2

No of mol of H2 liberated(n) = PV/RT

            P = 1 atm

            V = 75.5 ml = 0.0755 L

        R = 0.0821 l.atm.k-1.mol-1

        T = 273.15 k   (at STP)

          

              = 1*0.0755/(0.0821*273.15)

              = 0.00337 mol

No of mol of K reacted = 0.00337*2 = 0.00674 mol

mass of K reacted = 0.00674*39.1 = 0.263 g

5)

No of mol of CO2 reacted(n) = PV/RT = 1*15/(0.0821*273.15) = 0.67 mol

No of mol of NH3 taken =   1*10/(0.0821*273.15) = 0.446 mol

from equation, 1 mol CO2 = 1 mol NH3

so that, limiting reactant = NH3

no of mol of NaHCO3 formed = 0.446mol

mass of NaHCO3 formed = 0.446*84 = 37.5 g