DATA GIVEN:

Mass of BaO₂ = 1.45g
Volume BaO₂ = 25.5 mole
Mass of HCl 0.0277g/dm3
Asked:
a) To find the mass of H₂O₂
b) To find the mass of excess remain
Equation for the reaction
BaO_2(s) + 2HCl_(aq) → H₂O_2(aq) + BaCl_2(aq)

To find mass of HCl
From: ConC = mass /Volume
= ConC × volume
= 0.0277g/mole × 25.5 mole
= 0.70635g
To convert 0.70635g into mole
Mole = mass /Molar mass of HCl
= 0.70635g /36.5g/mole
= 0.01935 mole for HCl
Mole ratio for HCl is 0.00967

To find the mole for BaO₂
= Mass /Molar mass
= 1.45g /137g/mole
= 0.0085
Mole ratio = mole reacted /Mole coefficient
= 0.0085 mole /1 mole
= 0.0085

The limiting is HCl with 0.01935 mole the excess is BaO₂

a) To find the mass of H₂O₂

a) To find mass of H₂O₂ mole equivalent between HCl and H₂O₂
2 moles of HCl Ξ 1 mole H₂O₂
0.01935 mole HCl Ξ ??

= 0.01935 mole × 1 mole/2 mole
= 0.009675 mole for H₂O₂
Mass = mole × molar mass of H₂O₂
= 0.009675 mole × 34g/mole
= 0.32895g
0.32895g of hydrogen peroxide (H₂O₂) was produced

b) To find the mass of excess remain

b) Mass of excess of BaO₂ mole equivalent between BaO₂ and HCl
1 mole of BaO₂ Ξ 2 moles HCl
?? Ξ 0.01058 HCl

= 1 × 0.01058 /2
= 0.00529 mole of BaO₂

To convert 0.00529 into mass,
Mass = 0.00529 × 169
= 0.89401g BaO₂
Mass remain = 1.45g – 0.894g
= 0.556g
The mass remain of BaO₂ was 0.556g