Question

p13.16)Container A holds 722 mL of ideal gas at 2.80 atm. Container B holds 154 mL of ideal gas at 4.40 atm. If the gases are allowed to mix together, what is the partial pressure of each gas in the total volume? (atm)

p13.25) What concentration of ClO3– results when 813 mL of 0.673 M AgClO3 is mixed with 917 mL of 0.619 M Mn(ClO3)2? (M)

Answer 1

Given, [AgC_10_3] = 0.673 M, V = 813 ml or 0.813 L & [Mn (ClO_3)_2] = 0.619 M, V = 917 ml or 0.917 L rightarrow[We need to find moles of each species] (n) moles = molarity times volume (L) 1) n_AgClO_3 = M times V = (0.673) (0.813) - 0.547 moles Similarly n_Mn(uO_3)_2 = M times V = (0.619) (0.917) = 0.568 moles Now Ag CLO_3rightarrow Ag^+ + ClO^-_3 = > [n_ClO^-_3 = n_Ag ClO_3 = 0.547 moles [1 mole of AgClO_3 gives 1 mole of ClO^-_3] Refer stoichiometric coefficient, n_AgClO_3L n_ClO^-_3 = 1: 1 \r\n Similarly, Mn (ClO_3)_2 rightarrow Mn^+2_(ag) + 2 ClO)^-_3 = > n_ClO^-_3 (2) = 2 n_Mn (ClO_3)_2 = 2 times 0.568 = 1.136 moles Therefore Total moles of ClO_3, n_Total = > n_Total = n_ClO^-_3 (1) + n_ClO^-_3 (2) = 0.547 + 1.136 [n_Total/ClO^-_3 = 1.683 moles] Therefore final concentration of ClO^-_3 = > [ClO^-_3]_f = n_total/(V_total = V_1 + V_2)