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6 March, 12:57

In the reaction of silver nitrate with sodium chloride, how many grams of silver chloride will be produced from 100. g of silver nitrate when it is mixed with an excess of sodium chloride? The equation for the reaction is below. AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

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  1. 6 March, 15:16
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    85g

    Explanation:

    Firstly, from the balanced equation, we can see that one mole of silver nitrate yielded one mole of silver chloride. This is the theoretical relation. We need to get the actual relation. To do this, we will first need to get the number of moles of silver nitrate.

    The number of moles is the mass of silver nitrate divided by the molar mass of silver nitrate. The molar mass of silver nitrate is 108 + 14 + 3 (16) = 108 + 14 + 48 = 170g/mol

    The number of moles is thus 100/170 = 0.59 moles

    Since the mole ratio is 1 to 1, the number of moles of silver chloride produced too is 0.59 moles.

    To get the mass of silver chloride produced, we simply multiply the number of moles of silver chloride by the molar mass of silver chloride. The molar mass of silver chloride is 108 + 35.5 = 143.5g/mol

    The mass is thus = 143.5 * 0.59 = 84.67g
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