How many grams of silver sulphide are formed when 1.90 g of silver reacts with 0.280 g
of hydrogen sulphide and 0.160 g of oxygen?
Answer
first step: write the reaction equation: 4Ag + O2 + 2H2S = 2Ag2S + 2H2O
next, express each amount in moles
You can see that you will need 4 moles of Ag and 2 moles of H2S for each mole of O2
So check your amounts to see which will run out first. Then you can see how many moles of Ag2S will be formed, and express that in grams.
To determine the grams of silver sulfide formed, we need to first write out the balanced chemical equation for the reaction between silver, hydrogen sulfide, and oxygen:
2Ag + H2S + O2 -> Ag2S + H2O
From the balanced equation, we can see that 2 moles of silver react with 1 mole of hydrogen sulfide and 1 mole of oxygen to produce 1 mole of silver sulfide.
Next, let's calculate the number of moles of each substance:
Molar mass of Ag: 107.87 g/mol
Molar mass of H2S: 34.08 g/mol
Molar mass of O2: 32.00 g/mol
Molar mass of Ag2S: 247.8 g/mol
Number of moles of Ag: 1.90 g / 107.87 g/mol ≈ 0.0176 mol
Number of moles of H2S: 0.280 g / 34.08 g/mol ≈ 0.0082 mol
Number of moles of O2: 0.160 g / 32.00 g/mol ≈ 0.0050 mol
From the balanced equation, we can determine that the limiting reactant is the compound that we have the smallest number of moles. In this case, the limiting reactant is oxygen with 0.0050 mol.
Since 1 mole of O2 produces 1 mole of Ag2S, we can conclude that 0.0050 mol of O2 will produce 0.0050 mol of Ag2S.
Finally, let's calculate the mass of Ag2S:
Mass of Ag2S = number of moles of Ag2S x molar mass of Ag2S
Mass of Ag2S = 0.0050 mol x 247.8 g/mol ≈ 1.24 g
Therefore, approximately 1.24 grams of silver sulfide are formed when 1.90 g of silver reacts with 0.280 g of hydrogen sulfide and 0.160 g of oxygen.
To determine the number of grams of silver sulfide formed, we first have to balance the chemical equation for the reaction between silver, hydrogen sulfide, and oxygen.
The balanced chemical equation for the reaction is:
2 Ag + H2S + O2 -> Ag2S + H2O
From the balanced equation, we can see that 2 moles of silver (Ag) react with 1 mole of hydrogen sulfide (H2S) and 1 mole of oxygen (O2) to form 1 mole of silver sulfide (Ag2S) and 1 mole of water (H2O).
To determine the number of moles of each reactant, we need to convert the given masses to moles using their respective molar masses.
The molar mass of silver (Ag) is 107.87 g/mol.
The molar mass of hydrogen sulfide (H2S) is 34.08 g/mol.
The molar mass of oxygen (O2) is 32.00 g/mol.
1. Convert the mass of silver (Ag) to moles:
moles of silver (Ag) = mass (Ag) / molar mass (Ag)
moles of silver (Ag) = 1.90 g / 107.87 g/mol
moles of silver (Ag) = 0.0176 mol
2. Convert the mass of hydrogen sulfide (H2S) to moles:
moles of hydrogen sulfide (H2S) = mass (H2S) / molar mass (H2S)
moles of hydrogen sulfide (H2S) = 0.280 g / 34.08 g/mol
moles of hydrogen sulfide (H2S) = 0.0082 mol
3. Convert the mass of oxygen (O2) to moles:
moles of oxygen (O2) = mass (O2) / molar mass (O2)
moles of oxygen (O2) = 0.160 g / 32.00 g/mol
moles of oxygen (O2) = 0.005 mol
Now, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed. To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
According to the balanced equation, the stoichiometric ratio between silver (Ag), hydrogen sulfide (H2S), and oxygen (O2) is 2:1:1. Therefore, 2 moles of silver (Ag) react with 1 mole of hydrogen sulfide (H2S) and 1 mole of oxygen (O2).
From the moles calculated above:
The moles ratio of Ag:H2S:O2 is 0.0176 : 0.0082 : 0.005
Comparing these ratios, we can see that the mole ratio of Ag:H2S(0.0082) is almost double the mole ratio between Ag:O2(0.005). Therefore, hydrogen sulfide (H2S) is the limiting reactant.
To calculate the number of moles of silver sulfide (Ag2S) formed, we use the moles of limiting reactant (H2S).
The mole ratio between silver sulfide (Ag2S) and hydrogen sulfide (H2S) is 1:1. Therefore, the number of moles of silver sulfide (Ag2S) formed is equal to the number of moles of hydrogen sulfide (H2S).
moles of silver sulfide (Ag2S) = moles of hydrogen sulfide (H2S) = 0.0082 mol
Finally, we convert the number of moles of silver sulfide (Ag2S) to grams using its molar mass.
The molar mass of silver sulfide (Ag2S) is 247.80 g/mol.
grams of silver sulfide (Ag2S) = moles of silver sulfide (Ag2S) x molar mass (Ag2S)
grams of silver sulfide (Ag2S) = 0.0082 mol x 247.80 g/mol
grams of silver sulfide (Ag2S) = 2.03 g
Therefore, when 1.90 g of silver reacts with 0.280 g of hydrogen sulfide and 0.160 g of oxygen, 2.03 g of silver sulfide is formed.