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

Stoichiometric Calculations

The minimum q...

Question

The minimum quantity in grams of $H_{2} S$ needed to precipitate $63.5 g$ of ${C u}^{2 +}$ will be nearly:

63.5 g

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

31.75 g

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

34 g

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

20 g

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

Open in App

Solution

The correct option is D $34 g$
${C u}^{+ 2} + H_{2} S ⟶ C u S + H_{2}$
$1$ mole of ${C u}^{2 +}$ requires $1$ mole of $H_{2} S$
$∴ \frac{63.5}{63.5}$ moles of ${C u}^{2 +}$ requires $\frac{63.5}{63.5}$ moles of $H_{2} S$
i.e. $1$ mole of $H_{2} S = 1 \times 34 g$
$= 34 g$

Suggest Corrections

Similar questions

H_{2} S

63.5 g

C u^{2 +}

H_{2} S

63.5 g

C u^{2 +}

C u = 63.5 g/mol

C u^{2 +} + H_{2} S \to C u S + 2 H^{+}

H_{2} S

63.5

{C u}^{2 +}

{C u}^{+ 2} + H_{2} S \to C u S + H_{2}

H_{2} S

63.5 g

C u^{2 +}

H_{2} S

63.5 g

C u^{2 +}

C u = 63.5 g/mol

C u^{2 +} + H_{2} S \to C u S + 2 H^{+}

Join BYJU'S Learning Program