Question

Calcute the volume of hydrogen gas when 130 gm of zinc is combined with excess amount of hcl

Answer 1

Zn + 2HCl —> ZnCl2 + H2

to determine the limiting reactant, we first find the moles Of each reactant by dividing grams present by its molar mass:

8.89 g Zn x (1 mole Zn/65 g Zn)= 0.137 moles Zn

3.25 g HCl x (1 mole HCl/36 g HCl) = 0.0903 moles HCl

we then convert moles of one to moles of the other using their ratio from the balanced equation. This tells us how much of the other reactant we would theoretically need to fully react with whats present:

0.137 moles zn x (2 mole HCl/1 mole zn) = 0.274 moles HCl

since we only have 0.0903 moles hcl (compared to the 0.274 moles we need), we will run out of hcl before all of the zn is fully reacted. Therefore hcl is limiting.

to solve the rest of the problem, we need to find the theoretical yield. This is done using the mole ratio again, this time for limiting reactant to product:

0.0903 moles hcl x (1 mole zncl2/ 2 mole hcl) = 0.04515 moles zncl2

then we convert that value to mass of zncl2 that can be formed by multiplying the theoretical moles by its molar mass:

0.04515 moles zncl2 x (136 g/1 mole zncl) = 6.14 g zncl2

if the actual yield is 85%, the actual yield can be calculated by rearranging the equation for percent yield:

(actual/theoretica) x 100 = 85%

actual/6.14 = 0.85

actual = 5.23 g