What mass of manganese(ii) chloride must react with sulfuric acid to release 52.5 ml of hydrogen chloride gas at stp?
mncl2(s)+h2so4(aq)→mnso4(aq)+2hcl(g )

The balanced chemical reaction is written as:

MnCl2(s)+H2SO4(aq)→MnSO4(aq)+2HCl(g)

We are given the amount of hydrochloric acid to be produced in the reaction. This value will be the starting point for the calculations.We do as follows:

.0525 L HCl ( 1 mol / 22.4 L ) ( 1 mol MnCl2 / 2 mol HCl ) (125.84 g / 1 mol ) = 0.15 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 !

So what you would do is go through and add 3 teaspoons(tsp) 6 times because of the 6 tablespoons(tbsp). which will give you an answer of 18. another way is to multiply 3x6=18.