What mass of silver chloride can be produced from 1.88 L of a 0.139 M solution of silver nitrate?

Answer 1

Since the four named compounds are the only reactants and products, this question can be answered from the law of conservation of mass: The amount of silver nitrate must be 14.35 + 8.5 - 5.85 or 17.0 grams.

Answer 2

The formula for silver nitrate is AgNO3, and it has a gram formula mass of 169.87. The formula shows that each formula mass contains one silver atom, which has a gram Atomic Mass of 107.8688. Therefore, the mass fraction of silver in silver nitrate is 107.869/169.87.

1.33 L of a 0.129 M solution of silver nitrate, by the definition of molarity, contains (0.129)(1.33) or 0.17157 formula masses of silver nitrate, or 29.145 g of silver nitrate, or 18.507 g of silver.

The formula for silver chloride is AgCl, showing that it also has one silver atom per formula mass. The formula mass of silver chloride is 143.32. Therefore the ratio of mass of silver chloride to that of silver is 143.32/107.869, so that the mass of silver chloride that can be produced from the 29.145 g of silver present in the given amount of silver nitrate solution is 29.145(143.32/107.869) or 38.7 g of silver chloride, to the justified number of significant digits.

Answer 3

By definition, a 0.151 molar solution contains 0.151 moles of silver nitrate per liter of solution. Therefore, 1.20 L will contain 1.20(0.151) or 0.181 moles of silver nitrate. The formula units of both silver nitrate and silver chloride contain one atom of silver each, and the solubility product constant of silver chloride is sufficiently small that as many moles of silver chloride as of silver nitrate used can be completely converted to silver chloride, for which the gram formula unit mass is 143.22. Therefore, 143.22(0.181) or 26.0 grams of silver chloride, to the justified number of significant digits, can be formed

Answer 4

first write your reaction:

AgNO3 + NaCl >> AgCl +NaNO3

.177mol/L * 1.70L*(1mol AgCl/1mol AgNO3)*143 g/mol = 43.0287

Answer 5

1.66 L x 0.233 moles/L x 143 g/mol = 55.3 grams

Answer 6

The mass of silver chloride produced from 1,88 L of a 0,139 M solution of silver nitrate is 37,45 g.

Answer 7

approx 14.3g will be produced

Answer 8

116g AgCl