if a 0.625g sample of magnesium chloride is dissolved in water and then precipitated with a silver nitrate solution,how many grams of silver chloride precipitate would be collected?
To determine the number of grams of silver chloride precipitate that would be collected, we need to use stoichiometry and the molar masses of the compounds involved.
First, let's write the balanced equation for the reaction between magnesium chloride (MgCl2) and silver nitrate (AgNO3):
MgCl2 + 2AgNO3 → Mg(NO3)2 + 2AgCl
From the balanced equation, we can see that one mole of MgCl2 reacts with two moles of AgNO3 to form two moles of AgCl.
Next, calculate the number of moles of MgCl2:
moles of MgCl2 = mass of MgCl2 (in grams) / molar mass of MgCl2
The molar mass of MgCl2 is approximately 95.211 grams/mol.
moles of MgCl2 = 0.625 g / 95.211 g/mol
Now, since the reaction has a stoichiometric ratio of 1:2 between MgCl2 and AgCl, we know that the number of moles of AgCl will be twice the number of moles of MgCl2.
moles of AgCl = 2 * moles of MgCl2
Finally, calculate the mass of AgCl precipitate using the number of moles of AgCl and its molar mass. The molar mass of AgCl is approximately 143.321 grams/mol.
mass of AgCl = moles of AgCl * molar mass of AgCl
Now you have all the information needed to solve for the mass of AgCl precipitate. Substitute the values into the equation:
mass of AgCl = (2 * 0.625 g / 95.211 g/mol) * 143.321 g/mol
Calculate this expression to find the mass of AgCl precipitate.