Question

When sodium chloride reacts with concentrated sulfuric acid, hydrogen chloride gas (HCl) is one of the products. 

H2SO4 (l) + NaCl (s) → NaHSO4 (s) + HCl (g)

If hydrogen chloride gas is dissolved in water, a solution of hydrochloric acid is formed. How If we react 9,8gram H2SO4 with 73gram NaCl. Determine:

a. Limiting reactant                   [4]

b. How many grams remaining agent is it?                  [3]

c. How many grams NaHSO4 will be produced?                [3]

d. How many liters HCl will be produced if we measure in STP condition?      [3]

Answer 1

Moles of $H_2SO_4=\frac{9.8}{98.079}=0.1$

Moles of $NaCl=\frac{73}{58.44}=1.25$

Limiting reagent is $NaCl$

Mass of $NaHSO_4= 120.06×0.1=12.006g$

Mass of $HCl= 36.46×0.1=3.646g$