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Stoichiometric Calculations

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 $710 g$ of $K O_{2}$ having 60 % purity reacts.
(Molar mass of $K = 39 g/mol$ )

668 g

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339 g

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414 g

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467 g

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Solution

The correct option is C $414 g$
$4 K O_{2} (s) + 2 C O_{2} (g) \to 2 K_{2} C O_{3} (s) + 3 O_{2} (g)$
Molar mass of $K O_{2} = 71 g/mol$
Mass of pure $K O_{2}$ in the impure sample $= 710 \times 0.60 = 426 g$
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 $K_{2} C O_{3}$
$426 g$ will produce $= \frac{2 \times 138}{4 \times 71} \times 426 = 414 g$ of $K_{2} C O_{3}$

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Similar questions

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

2130 g

K O_{2}

K = 39 g/mol

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C O_{2}

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

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)

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48

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%)

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1 M

1.04 g

m L^{- 1}

K_{b}

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K = 39 g / m o l

C l = 35.5 g / m o l

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