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)

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

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this