a) The white precipitate formed in the reaction is silver halide (AgX), where X is the halide ion present in the salt MX.
b) The number of moles of silver halide precipitate produced can be calculated as follows:
moles of AgX = mass / molar mass = 5.63 g / (107.87 g/mol) = 0.052 moles
c) From the balanced chemical equation for the reaction, we know that 1 mole of MX reacts with 1 mole of AgNO3 to form 1 mole of AgX. Therefore, the number of moles of halide ion in the MX sample is also 0.052 moles.
mass of halide = moles of halide ion x molar mass of halide ion
Since we don't know which halide ion is present, we can calculate the masses for each possible option:
For chloride (Cl-): mass of halide = 0.052 moles x 35.45 g/mol = 1.84 g
For bromide (Br-): mass of halide = 0.052 moles x 79.90 g/mol = 4.16 g
For iodide (I-): mass of halide = 0.052 moles x 126.90 g/mol = 6.60 g
Therefore, the mass of halide present in the 2.3 g sample depends on the identity of the halide ion.
d) The percentage mass of halide in the MX sample can be calculated as follows:
% halide = (mass of halide / mass of MX) x 100%
Since we don't know the mass of MX, we cannot calculate this value.
e) The number of moles of metal M present in the halide salt MX can be calculated by subtracting the number of moles of halide ion from the total number of moles in the sample:
moles of metal M = total moles - moles of halide ion
total moles = mass / molar mass = 2.3 g / molar mass of MX
We cannot calculate the molar mass of MX without knowing the identity of the metal and halide ion.
f) The mass of metal M present in the halide salt MX can be calculated by multiplying the number of moles of metal M by its molar mass. However, we do not have enough information to calculate this value.
g) We cannot calculate the molar mass of the metal M present in the halide salt without knowing the identity of the metal and halide ion.