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Stoichiometric Calculations

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Question

How much sample of $B a C l_{2} .2 H_{2} O$ should be taken to get 0.75 mole of $B a^{2 +}$ ions?
$(B a = 137; C l = 35.5; H = 1; O = 16)$

283 g

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183 g

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383 g

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180 g

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Solution

The correct option is B 183 g
Molar mass of $H_{2} O = (137) + 2 (25.5) + 2 (18)$
$= 244$ grams.
1 mole gives 1 moles $B a^{2 +}$ ions.
0.75 moles gives 0.75 mole $B a^{2 +}$ ions.
1 mole of $B a C l_{2} .2 H_{2} O = 244$ grams.
0.75 mole of $B a C l_{2} .2 H_{2} O = (244) (0.75) = 183$ grams
$183$ grams of $B a C l_{2} .2 H_{2} O$ is required.

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Similar questions

Q. How much sample of

B a C l_{2} . 2 H_{2} O

will you take, if you need 0.75 mol of

B a^{2 +}

ions?

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

Q. The number of moles of water in 976 g

{B a C l}_{2} 2 H_{2} O

is:

(atomic masses of

H, O, C l,

and

B a

are 1, 16, 35, 5, and 137 respectively)

Q. The amount of

B a S O_{4}

formed upon mixing

100

mL of

20.8 %

B a C l_{2}

solution with

50

mL of

9.8 %

H_{2} S O_{4}

solution will be:
[Ba

= 137

, Cl

= 35.5

, S

= 32

, H

= 1

and O

= 16

Q. 100 mL of 20.8%

B a C l_{2}

solution and 50 mL of 9.8%

H_{2} S O_{4}

solution will form

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

Q. The amount of

B a S O_{4}

formed upon mixing

100

ml of

20.8 %

B a C l_{2}

solution with

50

ml of

9.8 %

H_{2} S O_{4}

solution will be:

(Given that molecular weight of

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

and

O = 16

g/mol)

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How much sample of BaCl2.2H2O should be taken to get 0.75 mole of Ba2+ ions?(Ba=137; Cl=35.5; H=1; O=16)

The correct option is B 183 gMolar mass of H2O=(137)+2(25.5)+2(18)=244 grams.1 mole gives 1 moles Ba2+ ions.0.75 moles gives 0.75 mole Ba2+ ions.1 mole of BaCl2.2H2O=244 grams.0.75 mole of BaCl2.2H2O=(244)(0.75)=183 grams183 grams of BaCl2.2H2O is required.

How much sample of $B a C l_{2} .2 H_{2} O$ should be taken to get 0.75 mole of $B a^{2 +}$ ions?
$(B a = 137; C l = 35.5; H = 1; O = 16)$