You visited us 1 times! Enjoying our articles? Unlock Full Access!

Hydrogen Peroxide

For the balan...

Question

For the balanced chemical reaction,
$4 K O_{2} (s) + 2 C O_{2} (g) \to 2 K_{2} C O_{3} (s) + 3 O_{2} (g)$
Determine the weight of solid potassium carbonate obtained when $2130 g$ of $K O_{2}$ having 50% purity reacts.
(Molar mass of $K = 39 g/mol$ )

414 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1035 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1242 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

808 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 1035 g
$4 K O_{2} (s) + 2 C O_{2} (g) \to 2 K_{2} C O_{3} (s) + 3 O_{2} (g)$
Molar mass of $K O_{2} = 71 g/mol$
Mass of pure $K O_{2}$ in the impure sample $= 2130 \times 0.50 = 1065 g$
As per the stoichiometry of the reaction,
4 mol of $K O_{2}$ reacts to form 2 mol of $K_{2} C O_{3}$
$(4 \times 71) g$ of $K O_{2}$ reacts to form $(2 \times 138) g$ of $K_{2} C O_{3}$
$1065 g$ will produce $= \frac{2 \times 138}{4 \times 71} \times 1065 = 1035 g$ of $K_{2} C O_{3}$

Suggest Corrections

Similar questions

4 K O_{2} (s) + 2 C O_{2} (g) \to 2 K_{2} C O_{3} (s) + 3 O_{2} (g)

710 g

K O_{2}

K = 39 g/mol

K O_{2}

C O_{2}

K O_{2} + C O_{2} \to K_{2} C O_{3} + O_{2}

C u S O_{4} .5 H_{2} O

K C N

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O (Yield = 100%)

C u S O_{4} + 2 K C N \to C u (C N)_{2} + K_{2} S O_{4} (Yield = 80%)

2 C u (C N)_{2} \to 2 C u C N + (C N)_{2} (Yield = 60%)

2 C u C N + 6 K C N \to 2 K_{3} [C u (C N)_{4}] (Yield = 50%)

C u S O_{4} .5 H_{2} O

K O_{2}

C O_{2}

H_{2} O

4 K O_{2} (s) + 2 H_{2} O (l) \to 3 O_{2} (g) + 4 K O H (s)

C O_{2}

K O H_{s w} + C O_{2} (g) \to K H C O_{3} (s)

K O_{2}

48

Join BYJU'S Learning Program