An anhydrous metal chloride, MCln, (1.500 g) is dissolved in water and the solution reacts with silver nitrate, requiring 47.34cm^3 of 0.5000M solution, to precipitate all the chloride as silver chloride. Identify M.
(0.4734 dm^3)(0.5000 mol/dm^3) = ____moles Ag+
Moles Cl- = moles Ag+
Grams of Cl- = (35.453 g/mol)(moles Cl-)
Grams of metal, M, = 1500g - grams Cl-
From the grams of M and grams of Cl- find the % composition of the sample.
Find the % composition of some chlorides: FeCl2, FeCl3, CuCl2, CaCl2, etc. until you find the one that matches the % composition of your sample.
To determine the identity of M, we can use stoichiometry and the volume and concentration of the silver nitrate solution.
First, we need to determine the number of moles of silver chloride that reacted with the given volume and concentration of silver nitrate. This can be calculated using the equation:
moles of AgCl = volume of AgNO3 solution (L) x concentration of AgNO3 (mol/L)
Converting the volume of the AgNO3 solution to liters:
volume of AgNO3 solution (L) = 47.34 cm^3 * (1 L / 1000 cm^3) = 0.04734 L
Now, calculating the moles of AgCl:
moles of AgCl = 0.04734 L * 0.5000 mol/L = 0.02367 mol
From the balanced equation, we know that one mole of MCln reacts with one mole of AgCl. Therefore, the number of moles of MCln is equal to the number of moles of AgCl:
moles of MCln = 0.02367 mol
Now, we can calculate the molar mass of MCln using the mass given in the problem (1.500 g) and the number of moles:
molar mass of MCln = mass of MCln (g) / moles of MCln
molar mass of MCln = 1.500 g / 0.02367 mol
molar mass of MCln = 63.436 g/mol
So, the molar mass of MCln is approximately 63.436 g/mol.
To find the identity of the metal chloride, M, we can use the stoichiometry of the reaction between the metal chloride and silver nitrate. The balanced equation for the reaction is:
MCln + AgNO3 → AgCl + MNO3
From the equation, we can see that one mole of MCln reacts with one mole of AgNO3 to produce one mole of AgCl. We can use this information to calculate the moles of AgNO3 required to precipitate all the chloride.
First, we need to calculate the number of moles of AgNO3 used in the reaction. To do this, we use the formula:
moles = concentration × volume
Given that the concentration of the silver nitrate solution is 0.5000 M and the volume used is 47.34 cm^3 (which can be converted to liters by dividing by 1000), we can calculate the number of moles of AgNO3 used:
moles of AgNO3 = 0.5000 M × (47.34 cm^3 / 1000) L = 0.02367 moles
Since one mole of AgNO3 reacts with one mole of MCln, the number of moles of MCln can be calculated as:
moles of MCln = 0.02367 moles
Since we have 1.500 g of MCln, we can calculate its molar mass by dividing the mass by the number of moles:
molar mass of MCln = 1.500 g / 0.02367 moles = 63.47 g/mol
Therefore, the molar mass of the metal chloride, MCln, is approximately 63.47 g/mol.
To determine the identity of the metal, we need to refer to a periodic table and find an element with a molar mass close to 63.47 g/mol. In this case, the closest match is chlorine (Cl), which has a molar mass of 35.45 g/mol.
Therefore, the metal in the anhydrous metal chloride, MCln, is chlorine (Cl).