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

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

What does that mean you dum bum