How many mL of a 5 M solution of hydrochloric acid are needed to react with 31.5 g of zinc sulfide?

Question

...With the below equation taken into consideration:

ZnS (s) + 2 HCl (aq) → ZnCl2 (aq) + H2S (l)

Help is greatly appreciated!

Answer 1

I don't Know but hold you nose if you try it The product is H2S, the "rotten egg" gas

Answer 2

I don't have exact numbers of atomic weights, but the moles corresponding to 31.5 g is about 0.3 mole. Working with this estimate, the equation indicates that 0.6 moles of HCl is needed. Since each liter of 5M solution delivers 5 moles, to obtain 0.6 moles, 0.6/5 or 0.12 L of solution is needed. Again, this is an estimate, but follow the numbers and it will work out.

Answer 3

hay,. do the stochiometry
31.5 g x 1 mole ZnS / 97.4 g ZN s x 36.45 g HCl/ 2 moles HCl =6.19 g HCl
and then using the definition of Molarity 5 mol HCl/ 1 liter solution x 36.45 G hcl/ 1 mol HCl

182.25 Grams HCl/ liter solution (x)=6.19 grams

6.19 grams /182.25 grams per liter = .033 liters

you need like 3.3% of 1 liter of the solution so 1 liter x (.033)

.033 liters x 1000 ml/ 1 liter= 33 milliliters

ok g2g to work bye pal

Answer 4

1 mL = 1g. find moles of each product given by calculating molar mass of each compound. Use mole ratio and # of moles to determine the lilmiting reactant. convert to g, which is as same as mL.

Answer 5

molar mass of ZnS = ninety seven.5 gm/mol 31.5 gm/ninety seven.5gm/mol = 0.323 moles It demands 2 moles of HCl for each mole of ZnS, therefor you elect 2 x 0.323 moles of HCl or 0.646 moles of HCl xL (5 mol/L) = 0.646 moles X = 0.129 liters = 129 mls

Answer 6

hehe i had that equation in chemistry..im too lazy to get it though..trying to remember how to do it..ill let you know!