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3 years ago

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A 100.0 mL sample of a 0.200 M aqueous solution of K2CrO4 was added to 100.0 mL of a 0.100 M aqueous solution of BaCl2. The mixt

ure was stirred and the precipitate was collected, dried carefully, and weighed. How many grams of precipitate should be obtained? The reaction is shown below: K2CrO4(aq) + BaCl2(aq) ď‚® BaCrO4(s) + 2 KCl(aq) A. 2.05 g B. 2.53 g C. 5.07 g D. 6.16 g E. 7.60 g

1 answer:

stepan [7]3 years ago

5 0

The correct answer is 2.53 g of precipitate, BaCrO4.

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4 years ago

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The mass of oxygen produced has been 50 g. The moles of oxygen has been given by:

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The moles of oxygen produced has been 1.5625 mol.

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$\rm 1 \;mole\;O_2=2\;mole\;H_2O\\1.5625\;mol\;O_2=1.5625\;\times\;2\;mol\;H_2O\\1.5625\;mol\;O_2=3.125\;mol\;H_2O$

The moles of hydrogen decomposes has been 3.125 mol.

The mass of hydrogen decomposed has been given by:

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The mass of water decomposed to produce 50 g oxygen has been 56.28 g. Thus, option D is correct.

For more information about moles produced, refer to the link:

brainly.com/question/10606802

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