Chemistry (Gas)

Question

When hydrogen peroxide reacts with bleach (sodium hypochlorite), a very explosive reaction takes place. If 3 mL of a 0.45 mol/L solution of hydrogen peroxide reacts with 4 mL of a 0.32 mol/L solution of sodium hypochlorite, what mass of oxygen gas is produced? (6)

            H2O2 (aq)    +   NaOCl (aq)    → NaCl (aq) +H2O (aq)    + O2 (g) 

Answer 1

The equation is balanced:

H2O2 (aq)    +   NaOCl (aq)    → NaCl (aq) +H2O (aq)    + O2 (g) 

Find the limiting reactant:

3mL H2O2 available

4mL NaOCl available

3mL H2O2 needs _?_ mL NaOCl to react

(^{3mL H2O2}/₁)*(^0.45molH2O2/_1000mLH2O2)*(^1molNaOCl/_1molH2O2)*(^1000mLNaOCl/_0.32molNaOCl)= 4.22 mL NaOCl needed

4.22mLNaOCl needed > 4 mL NaOCl available

There's not enough NaOCl. NaOCl is the limiting reactant.

Use the amount of limiting reactant available to find how much product forms.

(^{4mL NaOCl}/₁)*(^0.32molNaOCl/_1000mLNaOCl)*(^1molO2/_1molNaOCl)*(^32gO2/_1molO2)=40.96 g O2