treatment of a 0.3203g sample of impure soduim chloride with an excess of AgNo3 resulted in the formation of 0.7331g of AgCl.
how many grams of sodium chloride were in the sample?
What is the weight percentage of NaCl in the sample?
See your post below.
To find the number of grams of sodium chloride in the sample, we can use the stoichiometry of the reaction between sodium chloride (NaCl) and silver nitrate (AgNO3).
The balanced equation for the reaction is:
NaCl + AgNO3 -> AgCl + NaNO3
From the equation, we can see that the molar ratio of NaCl to AgCl is 1:1, meaning that for every 1 mole of NaCl, 1 mole of AgCl is formed.
First, let's calculate the number of moles of AgCl formed:
Mass of AgCl = 0.7331g
Now, we need to convert the mass of AgCl to moles. The molar mass of AgCl is 143.32 g/mol.
Moles of AgCl = mass / molar mass = 0.7331g / 143.32 g/mol = approximately 0.005115 mol
Since the molar ratio of NaCl to AgCl is 1:1, the number of moles of NaCl is also 0.005115 mol.
To calculate the number of grams of NaCl present, we need to use the molar mass of NaCl, which is 58.44 g/mol.
Grams of NaCl = moles of NaCl * molar mass = 0.005115 mol * 58.44 g/mol = approximately 0.2988g
Therefore, the number of grams of sodium chloride in the sample is approximately 0.2988g.
To find the weight percentage of NaCl in the sample, we can use the formula:
Weight percentage of NaCl = (grams of NaCl / grams of sample) * 100%
The weight of the sample is given as 0.3203g.
Weight percentage of NaCl = (0.2988g / 0.3203g) * 100% = approximately 93.3%
Therefore, the weight percentage of NaCl in the sample is approximately 93.3%.