what mass of manganese (ii) chloride must react with sulfuric acid to release 49.5 ml of hydrogen chloride gas at stp? mncl2(s) + h2so4(> mnso4(aq) + 2hcl(g)

The mass of the manganese II chloride or Mncl2 must react with the sulfuric acid to release the 49.5 mL of hydrogen chloride gas at STP;
(0.0495L HCl) / (22.414 L/mol) x (1 mol MnCl2 / 2 mol HCl) x (125.8440 g MnCl2/mol) = 0.1389595342 approximately 0.139 g MnCl2
The mass of the manganese II chloride or Mncl2 will be 0.139 g MnCl2

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !