How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride in the reaction #2AgNO_3 + BaCl_2 -> 2AgCl + Ba(NO_3)_2#?

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How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride in the reaction #2AgNO_3 + BaCl_2 -> 2AgCl + Ba(NO_3)_2#?

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Answer 1 · 2016-03-04T19:34:21

#m=4.2gAgCl#

Explanation:

The balanced equation is:

#2AgNO_3+BaCl_2->2AgCl+Ba(NO_3)_2#

To find the mass of silver chloride when reaction 5.0g of silver nitrate, we can use dimensional analysis:

#?gAgCl=5.0cancel(gAgNO_3)xx(1cancel(molAgNO_3))/(169.9cancel(gAgNO_3))xx(2cancel(molAgCl))/(2cancel(molAgNO_3))xx(143.3gAgCl)/(1cancel(molAgCl))=4.2gAgCl#

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How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride in the reaction #2AgNO_3 + BaCl_2 -> 2AgCl + Ba(NO_3)_2#?

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