2AgNO3 + BaCl2 --> 2AgCl + Ba(NO3)2
How many grams of sliver chloride are produced from 5.0g of sliver nitrate reacting with an excess of barium chloride?
mols AgNO3 = grams/molar mass
Convert mols AgNO3 to mols AgCl.
Convert mols AgCl to g. g = mols x molar mass.
fgagfa
To determine the number of grams of silver chloride produced, we first need to calculate the number of moles of silver nitrate (AgNO3) using its molar mass.
The molar mass of AgNO3 is:
AgNO3 = (1 Ag) + (1 N) + (3 O) = 108 g/mol
Now, we can calculate the number of moles of AgNO3:
Number of moles = Mass / Molar mass
Number of moles of AgNO3 = 5.0 g / 108 g/mol ≈ 0.0463 mol
From the balanced chemical equation:
2 AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2
We can see that the ratio of AgNO3 to AgCl is 2:2, which means that for every 2 moles of AgNO3, we get 2 moles of AgCl.
Therefore, the number of moles of AgCl produced is also 0.0463 mol.
Now, we can calculate the mass of AgCl using its molar mass.
The molar mass of AgCl is:
AgCl = (1 Ag) + (1 Cl) = 143 g/mol
Mass of AgCl = Number of moles × Molar mass
Mass of AgCl = 0.0463 mol × 143 g/mol ≈ 6.62 g
Therefore, approximately 6.62 grams of silver chloride are produced from 5.0 grams of silver nitrate reacting with an excess of barium chloride.