If the flask is filled with H2 and HCl the reaction Given in the question cannot proceed.
But if H2 and Cl2 is taken in the flask the above reaction can occur. Hence the concentration given for HCl taken in the flask is taken as concentration of Cl2 and the following calculaion is done. 0.1814mol of HCl is produced.
Suppose a 500. mL flask is filled with 1.3 mol of H2 and 0.10 mol of...
Suppose a 500. ml flask is filled with 0.10 mol of H, and 1.0 mol of HCl. The following reaction becomes possible: H2(g) +C1, (g) = 2HCl(g) The equilibrium constant K for this reaction is 4.69 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places. x 6 ?
Suppose a 250. ml flask is filled with 1.5 mol of Cl, and 1.3 mol of HCl. The following reaction becomes possible: H2(g) + Cl2(g) 2HCl (8) The equilibrium constant K for this reaction is 0.560 at the temperature of the flask. Calculate the equilibrium molarity of H. Round your answer to two decimal places. OM * 5 ?
Suppose a 250. mL flask is filled with 0.10 mol of Cl2 and 1.4 mol of HCl. The following reaction becomes possible: H2(g)+Cl2(g)=2HCl(g) The equilibrium constant for this reaction is 0.414 at the temperature of the flask. Calculate the equilibrium molarity of . Round your answer to two decimal places.
Suppose a 500. ml flask is filled with 1.2 mol of Cl, and 0.80 mol of HCl. The following reaction becomes possible: H2(g) + Cl2(g) = 2HCl (g) The equilibrium constant K for this reaction is 0.419 at the temperature of the flask. Calculate the equilibrium molarity of Cl. Round your answer to two decimal places. xs ?
Suppose a 500.mL flask is filled with 0.60mol of H2 and 0.70mol of Cl2. The following reaction becomes possible: H2(g) + Cl2(g) 2HCl(g) The equilibrium constant K for this reaction is 5.52 at the temperature of the flask. Calculate the equilibrium molarity of Cl2. Round your answer to two decimal places.
Suppose a 250. ml flask is filled with 1.6 mol of Cl, and 1.3 mol of HCI. The following reaction becomes possible: H2(g) + Cl2(g) → 2HCI (g) The equilibrium constant K for this reaction is 0.967 at the temperature of the flask. Calculate the equilibrium molarity of C1. Round your answer to two decimal places. OM I ?
Suppose a 500. ml flask is filled with 1.3 mol of NO, and 1.5 mol of NO. The following reaction becomes possible: NO2(g) +NO(g) + 2NO2(g) The equilibrium constant K for this reaction is 9.06 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. i M x 6 ?
Suppose a 500. mL flask is filled with 0.50 mol of H2 and 1.9 mol of HI. The following reaction becomes possible: H2(g)+12g2HIg The equilibrium constant K for this reaction is 4.94 at the temperature of the flask. Calculate the equilibrium molarity of H2. Round your answer to two decimal places.
Suppose a 500. mL flask is filled with 0.70 mol of H2 and 0.60 mol of HI. The following reaction becomes possible: -2H1(g) The equilibrium constant K for this reaction is 2.38 at the temperature of the flask. Calculate the equilibrium molarity of H2. Round your answer to two decimal places.
Suppose a 250. mL flask is filled with 0.30 mol of H2 and 1.3 mol of HI. The following reaction becomes possible: H2(8)+12)2HIg) The equilibrium constant K for this reaction is 0.254 at the temperature of the flask. Calculate the equilibrium molarity of H2. Round your answer to two decimal places. Ar