AgNO3+ BaCl2-> AgCl+ Ba(NO3)2
a. balence the equasion
b. How many grams of silver chloride(AgCl) are produced from 5.0g of silver nitrate (AgNO3) reacting with an excess of barium chloride(Ba Cl2)?
c. How Many grams of barium chloride (BaCl2) is necessary to react with 5.0 grams of silver nitrate (AgNO3)?
Can anyone help me with this????
Stoichiometry problems are easy to solve followoing a 4-step process.
Step 1. Write and balance the chemical equation.
2AgNO3 + BaCL2 ==> 2AgCl + Ba(NO3)2
Step 2. Convert what you have (in this case 5.0 g AgNO3) to mols remembering that mols = g/molar mass.
mols AgNO3 = 5.0/170 = 0.294 mols.
Step 3. Convert mols of what you have (in this case mols AgNO3) to mols of what you want (in this case AgCl). Use the coefficients in the balanced equation.
0.294 mols AgNO3 x (2 mols AgCl/2 mols AgNO3) = 0.294 x 2/2 = 0.294 x 1/1 = 0.294 mols AgCl. Note that the fraction is arranged so that the units of AgNO3 cancel and leave units of AgCl. That is how mols of AgNO3 are converted to mols AgCl. .
Step 4. Convert mols from step 3 (in this case 0.294 mols AgCl) to grams AgCl. Use the equation from Step 2 rearranged; i.e., mols=grams/molar mass so grams = mols x molar mass and g AgCl = mols AgCl x molar mass AgCl = 0.294 g AgCl x 143 g AgCl/mol AgCl = 4.21 g AgCl.
You will need to redo all of the above because I estimated the molar mass of AgNO3 at 170 and the molar mass of AgCl at 143. Recalculate those values and plug them into the appropriate spots to get an accurate answer. The 4.21 number will be close.
For part 2 of your problem, it is the same procedure. Post your work if you get stuck and need further assistance.
Remember this four-step procedure. It will work many many problems for you. A slight modification is necessary for limiting reagent problems.
i don't understand but this helps i hope it's right
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a. To balance the equation, we need to make sure that the number of atoms for each element on both sides of the equation is equal.
The balanced equation for the reaction is:
2AgNO3 + BaCl2 -> 2AgCl + Ba(NO3)2
b. To determine the grams of silver chloride (AgCl) produced, we need to use stoichiometry.
Step 1: Calculate the moles of silver nitrate (AgNO3) using the given mass and molar mass.
molar mass AgNO3 = 107.87 + 14.01 + 3(16.00) = 169.87 g/mol
moles AgNO3 = mass AgNO3 / molar mass AgNO3
moles AgNO3 = 5.0 g / 169.87 g/mol = 0.0294 mol AgNO3
Step 2: Use the balanced equation to find the mole ratio between AgNO3 and AgCl.
From the balanced equation: 2 mol AgNO3 produces 2 mol AgCl
moles AgCl = moles AgNO3 = 0.0294 mol
Step 3: Convert moles of AgCl to grams using the molar mass of AgCl.
molar mass AgCl = 107.87 + 35.45 = 143.32 g/mol
grams AgCl = moles AgCl x molar mass AgCl
grams AgCl = 0.0294 mol x 143.32 g/mol = 4.21 g AgCl
Therefore, 5.0 g of silver nitrate will produce 4.21 g of silver chloride.
c. To determine the grams of barium chloride (BaCl2) necessary to react with 5.0 g of silver nitrate (AgNO3), we follow the same steps as above.
Step 1: Calculate the moles of AgNO3 using the given mass and molar mass.
moles AgNO3 = 5.0 g / molar mass AgNO3
Step 2: Use the balanced equation to find the mole ratio between AgNO3 and BaCl2.
From the balanced equation: 2 mol AgNO3 requires 1 mol BaCl2
Step 3: Convert moles of BaCl2 to grams using the molar mass of BaCl2.
grams BaCl2 = moles BaCl2 x molar mass BaCl2
Note: The molar masses used in the calculations may differ from the ones provided in the explanation. Make sure to use the accurate molar masses for silver nitrate (AgNO3) and silver chloride (AgCl).