Solve the following stoichiometry problems a) What volume in liters of 0.1825 M

Question

Solve the following stoichiometry problems a) What volume in liters of 0.1825 M KOH is needed to titrate 25.45 mL of 0.2555 M H_2SO_4 in the following neutralization reaction? H_2SO_4 + 2 KOH rightarrow 2 H_20 + K_2SO_4 b) 175.0 grams of C_6H_12O_6 is burned in excess oxygen and 96.5 grams of water is produced. What is the percent yield of water? C_6H_12O_6+ 6 O_2 rightarrow 6 CO_2 + 6 H_2O c) What is the empirical formula of a compound that is 73.9 % Hg and 26.1 % CI? d) What mass in grams of NH_3 can be produced when 84.0 g N_2 reacts with 12.1 gH_2? N_2 + 3 H_2 rightarrow 2 NH_3 e) How many atoms of carbon are found in 50.0 grams of CH_4? f) A sample 25.0 mL of an NaCI solution is diluted to a final volume of 125 of 200 M NaCl. What was concentration of the original NaCl solution.

Explanation / Answer

9
a) M1V1/n1 = M2V2/n2

   (25.45*0.2555)/1 = (0.1825*v2)/2

V2 = volume of NaOH= 71.26 ml

b) from equation

   1 mole C6H12O6 = 6 mole H2O

No of mole of C6H12O6 = 175/180 = 0.972 mole

theroretical yield of H2O = 0.972*6 = 5.832 mole

practical yield of H2o = 96.5/18 = 5.36 mole

%yield = 5.36/5.832*100 = 91.9%

c) no of mole of N2= 84/28 = 3 MOLE

   no of mole of H2 = 12.1/2 = 6.05 MOLE

   limiting reactant = H2

   no of mole of NH3 produced = 6.05*2/3 = 4.03 mole

mass of NH3 = 4.03*17 = 68.51 g

D) no of atoms of c = 50/16*(6.023*10^23) = 1.88*10^24 atoms

e) not visible clearly

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