17g of AgNo3 is trated with 25g of HCl. what is the mass of AgCl formed and what is the limiting reagent
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To determine the mass of AgCl formed and the limiting reagent, we need to use stoichiometry and compare the moles of reactants.
First, let's calculate the moles of each compound:
Molar mass of AgNO3: 107.87 g/mol
Molar mass of HCl: 36.46 g/mol
Moles of AgNO3 = mass / molar mass
Moles of AgNO3 = 17 g / 107.87 g/mol = 0.1578 mol
Moles of HCl = mass / molar mass
Moles of HCl = 25 g / 36.46 g/mol = 0.685 mol
The balanced equation for the reaction between AgNO3 and HCl is:
AgNO3 + HCl -> AgCl + HNO3
From the balanced equation, we can see that the stoichiometric ratio between AgNO3 and AgCl is 1:1.
Since the moles of AgNO3 are 0.1578 mol, the moles of AgCl formed will also be 0.1578 mol.
Now let's calculate the mass of AgCl formed:
Mass of AgCl = moles of AgCl * molar mass of AgCl
Mass of AgCl = 0.1578 mol * 143.32 g/mol = 22.68 g
Therefore, the mass of AgCl formed is 22.68 g.
To determine the limiting reagent, we compare the moles of AgNO3 and HCl. The reactant that produces fewer moles of AgCl is the limiting reagent.
The stoichiometric ratio between AgNO3 and AgCl is 1:1.
Since the moles of AgCl formed are 0.1578 mol, the moles of AgNO3 required would also be 0.1578 mol.
Comparing the moles of AgNO3 (0.1578 mol) and HCl (0.685 mol), we can see that AgNO3 is the limiting reagent because it produces fewer moles of AgCl.
Therefore, AgNO3 is the limiting reagent.
To determine the mass of AgCl formed and identify the limiting reagent in this chemical reaction, we need to compare the stoichiometry of the reactants.
First, let's write and balance the chemical equation for the reaction between AgNO3 and HCl:
AgNO3 + HCl → AgCl + HNO3
The balanced equation shows that 1 mole of AgNO3 reacts with 1 mole of HCl to produce 1 mole of AgCl.
Next, we need to calculate the number of moles of each reactant. To do this, we use the formula:
moles = mass / molar mass
The molar mass of AgNO3 is calculated as follows:
Ag = 107.87 g/mol
N = 14.01 g/mol
O = 3 x 16.00 g/mol = 48.00 g/mol
Total molar mass of AgNO3 = 107.87 + 14.01 + 48.00 = 169.88 g/mol
Using the given mass of AgNO3, we can calculate the number of moles:
moles of AgNO3 = 17 g / 169.88 g/mol = 0.10 mol
Similarly, for HCl:
H = 1.01 g/mol
Cl = 35.45 g/mol
Total molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol
moles of HCl = 25 g / 36.46 g/mol = 0.69 mol
Now, we have the number of moles of each reactant. To determine the limiting reagent, we compare the mole ratio of AgNO3 to HCl in the balanced equation.
From the balanced equation, we see that the mole ratio is 1:1. This means that for every 1 mole of AgNO3, we need 1 mole of HCl.
Since we have 0.10 moles of AgNO3 and 0.69 moles of HCl, we have an excess of HCl. HCl is the limiting reagent.
To find the mass of AgCl formed, we use the mole ratio mentioned earlier. From the balanced equation, 1 mole of AgNO3 produces 1 mole of AgCl. Therefore, the mass of AgCl formed is equal to the molar mass of AgCl:
Ag = 107.87 g/mol
Cl = 35.45 g/mol
Total molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol
The mass of AgCl formed can be calculated as follows:
mass of AgCl = moles of AgCl x molar mass of AgCl
mass of AgCl = moles of AgNO3 x molar mass of AgCl
mass of AgCl = 0.10 mol x 143.32 g/mol = 14.33 g
Therefore, the mass of AgCl formed is 14.33 grams and the limiting reagent is HCl.