Question

Gaseous ammonia, NH_3, and hydrogen chloride, HCI, are known to react to produce ammonium chloride, NH_4CI according to the equation shown below. NH_4CI could be used as smelling salts (although I wouldn't recommend it) NH_3(g) + HCl(g) rightarrow NH_4C1 (s) If you are given 7.38 g of NH3 and 9.32 g of HCI, what mass of NH_4Cl will be produced A 20.0 mL sample of a Ba(OH)_2 solution was found to react with 28.60 mL of 0.125 M HCI. What is the concentration of the Ba(OH)_2 solution.Ba(OH)_2 + 2HC1 rightarrpow BaCl_2 + 2H_2O

Answer 1

No of moles of NH_3 = 7.38g times 1 mol NH_3/17.04 g_3 = 0.433mol NH_3 No of moles of HCL = 9.32g times 1 mol HCL/36.45g = 0.255 mol of HCL Equation: NH_3 rightarrow NH_4 CL No of moles of NH_4 CL produced from NH_4 = 0.433 mol NH_3 times 1 moles NH_4 CL/1 mol NH_3 = 0.433 mol NH_4 CL No of moles of NH_4 CL produced from HCL = 0.433 mol NH_3 times 1 moles NH_4 CL/1 mol HCL = 0.255 mol NH_4 CL Since HCL limits the produced = 0.255 mol Theoretical yield of NH_4 Cl = 0.255 mol NH_4 CL times 53.5 g/1 mol NH_4 Cl = 13.6g