# 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?

Mar 4, 2016

$m = 4.2 g A g C l$

#### Explanation:

The balanced equation is:

$2 A g N {O}_{3} + B a C {l}_{2} \to 2 A g C l + B a {\left(N {O}_{3}\right)}_{2}$

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