If 0.03 mol of zinc are added to 0.52 mol of $HCl$ , the moles of $H_{2}$ formed are:

0.52

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

0.30

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

0.26

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

0.60

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

Open in App

Solution

The correct option is C $0.26$
$Z n + 2 H C l ⟶ Z n C l_{2} + H_{2}$
$0.03$ $0.52$
$0.46$ $0.03$ $0.03$
So, $0.03$ $m o l e s$ of $H_{2}$ are formed as zinc is the limiting reactant.

Suggest Corrections

Similar questions

Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)

Z n

H C l

H_{2}

H C l

Z n + 2 H C l \to Z n C l_{2} + H_{2}

If 2 moles of sugar are dissolved in 1000g of water, the elevation in boiling point of the resulting solution is: (K_b = 0.52 K kg mol^-1):

{100.15}^{o} C .

K_{f}

K_{b}

m o l^{- 1}

m o l^{- 1}

Join BYJU'S Learning Program