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

Stoichiometric Calculations

50ml solution...

Question

$50 m l$ solution of $B a C l_{2}$ (20.8 % w/v) and $100 m l$ solution of $H_{2} S O_{4}$ (9.8 %w/v) are mixed then maximum mass of $B a S O_{4}$ formed is:
$[B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l]$

Open in App

Solution

$B a C l_{2}$ + $H_{2} S O_{4} \to B a S O_{4} + 2 H C l$
Mass of $B a C l_{2}$ = $\frac{20.8}{100} \times 50$ g
=10.4g
moles of $B a C l_{2}$ = $\frac{10.4}{208}$ = 0.05 moles
Mass of $H_{2} S O_{4}$ = $\frac{9.8}{100} \times 100$ =9.8g
Moles of $H_{2} S O_{4}$ = $\frac{9.8}{9.8}$ =0/.1 moles
Therefore, $B a C l_{2}$ is a limiting reagent
Therefore , maximu moles of $B a S O_{4}$ formed is 0.05 moles
Maximum mass of $B a S O_{4}$ formed=0.05 x233g
=11.65g
Therefore maximum mass of $B a S O_{4}$ formed is 11.65 g

Suggest Corrections

Similar questions

{BaCl}_{2}

H_{2} {SO}_{4}

{BaSO}_{4}

[{BaCl}_{2} {+ H}_{2} {SO}_{4} \to {BaSO}_{4} + 2HCl]

B a C l_{2}

H_{2} S O_{4}

B a S O_{4}

(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

B a C l_{2}

H_{2} S O_{4}

C l^{-}

B a C l_{2} (a q)

H_{2} S O_{4}

B a S O_{4}

B a S O_{4}

100

20.8 %

B a C l_{2}

50

9.8 %

H_{2} S O_{4}

B a = 137, C l = 35.5, S = 32, H = 1

O = 16

Join BYJU'S Learning Program