Silver reacts with hydrogen sulphide gas, and oxygen according to the reaction:
4Ag(s) + 2H2S(g) + O2(g) → 2Ag2S(s)+ 2H2O(g)
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?
This is a limiting reagent problem (LR) , and a difficult one at that, and the only way I know that you can be SURE you have it correct is to do it the long way. Basically that is to determine how many grams Ag2S formed from each of the reactants and to pick the smallest one. I hope oobleck shows you the short way.
4Ag(s) + 2H2S(g) + O2(g) → 2Ag2S(s)+ 2H2O(g)
.1.90 g.....0.280 g.....0.160 g
mols Ag = 1.90 g/107.9 = 0.0176
mols H2S = 0.280/34 = 0.00823
mols O2 = 0.160/32 = 0.005
Using the coefficients convert each reactant, by itself, to moles of product of Ag2S.
For Ag: 0.0176 moles Ag x (2 mols Ag2S/4 mols Ag) = 0.0088
For H2S: 0.00823 mols H2S x (2 moles Ag2S/2 mols H2S) = 0.00823
For O2: 0.005 mols O2 x (2 moles Ag2S/1 mol O2) = 0.01
The smallest number of moles Ag2S formed is from the H2S; therefore, 0.00823 mols Ag2S will be formed. Convert that to grams. g = mols x molar mass. Post your work if you get stuck.
Well, let's do some math to find out how many grams of silver sulphide are formed.
Based on the balanced equation, we can see that the ratio of Ag to Ag2S is 4:2, or simply 2:1.
Since we have 1.90 g of silver, we will divide it by 2 to get the mass of silver sulphide formed.
So, 1.90 g of silver would produce 0.95 g of silver sulphide.
Now, considering the mass of hydrogen sulphide, we have 0.280 g. Looking at the ratio in the balanced equation, we see that 2 moles of H2S react to form 2 moles of Ag2S. From here, we can convert the grams of H2S to moles and then to grams of Ag2S.
Using the molar mass of H2S (32.07 g/mol), we find that 0.280 g of H2S is equal to 0.00873 moles.
Since the ratio of H2S to Ag2S is 2:2, or 1:1, we can say that 0.00873 moles of H2S will produce 0.00873 moles of Ag2S.
Finally, let's look at the oxygen. We have 0.160 g of oxygen, which can be converted to moles using the molar mass of O2 (32.00 g/mol). This gives us 0.005 moles of O2.
Now, comparing the amounts of H2S and O2, we see that there is less oxygen than H2S. This means that oxygen is the limiting reactant in this case, and we only need to consider the amount of moles of Ag2S produced from the oxygen.
Since we have 0.005 moles of O2, and the ratio of O2 to Ag2S is 1:1, we can say that 0.005 moles of O2 will produce 0.005 moles of Ag2S.
Converting this back to grams, we have 0.005 moles of Ag2S, which is equal to 1.21 g of silver sulphide.
Therefore, the amount of silver sulphide formed when 1.90 g of silver reacts with 0.280 g of hydrogen sulphide and 0.160 g of oxygen is approximately 1.21 grams.
Keep in mind, though, that this answer is based on stoichiometry and assumes that the reaction proceeds completely, without any side reactions or losses.
To find the grams of silver sulphide formed, we need to calculate the limiting reactant first. The limiting reactant is the one that determines the maximum amount of product that can be formed.
Let's start by calculating the moles of each reactant:
Molar mass of Ag = 107.87 g/mol
Molar mass of H2S = 34.08 g/mol
Molar mass of O2 = 32.00 g/mol
Moles of silver (Ag):
moles of Ag = mass of Ag / molar mass of Ag
moles of Ag = 1.90 g / 107.87 g/mol
moles of Ag = 0.01764 mol
Moles of hydrogen sulphide (H2S):
moles of H2S = mass of H2S / molar mass of H2S
moles of H2S = 0.280 g / 34.08 g/mol
moles of H2S = 0.00821 mol
Moles of oxygen (O2):
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 0.160 g / 32.00 g/mol
moles of O2 = 0.005 mol
Based on the balanced equation, the ratio of Ag to Ag2S is 4:2, so the moles of Ag2S formed will be half of the moles of Ag.
Moles of Ag2S = 0.01764 mol / 2
Moles of Ag2S = 0.00882 mol
Now, let's calculate the mass of silver sulphide:
Mass of silver sulphide (Ag2S) = Moles of Ag2S * molar mass of Ag2S
Mass of Ag2S = 0.00882 mol * (2 * 107.87 g/mol)
Mass of Ag2S = 1.88 g
Therefore, 1.88 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.
To find the number of grams of silver sulphide formed when 1.90 g of silver reacts with 0.280 g of hydrogen sulfide and 0.160 g of oxygen, we first need to determine the limiting reactant.
1. Calculate the moles of each reactant:
Moles of silver (Ag) = mass of silver / molar mass of silver
= 1.90 g / 107.9 g/mol (molar mass of Ag)
= 0.0176 mol
Moles of hydrogen sulfide (H2S) = mass of H2S / molar mass of H2S
= 0.280 g / 34.1 g/mol (molar mass of H2S)
= 0.0082 mol
Moles of oxygen (O2) = mass of oxygen / molar mass of oxygen
= 0.160 g / 32.0 g/mol (molar mass of O2)
= 0.0050 mol
2. Determine the ratio of moles between the reactants and the product:
From the balanced chemical equation, we can see that the ratio between Ag, H2S, and O2 to Ag2S is 4:2:1.
Therefore, the ratio of moles between Ag and Ag2S is 4:2.
3. Find the limiting reactant:
To find the limiting reactant, we need to compare the ratios of Ag and H2S to Ag2S.
The ratio of Ag to Ag2S is 4:2 = 2:1.
The ratio of H2S to Ag2S is 2:1.
Since the ratio of Ag to Ag2S is the same as the ratio of H2S to Ag2S, we can conclude that the limiting reactant is H2S.
4. Calculate the moles of Ag2S formed using the limiting reactant (H2S):
From the balanced equation, we can see that the mole ratio between H2S and Ag2S is 2:2.
Therefore, the moles of Ag2S formed = 0.0082 mol (moles of H2S) × (2 mol Ag2S / 2 mol H2S)
= 0.0082 mol
5. Calculate the mass of Ag2S formed:
Mass of Ag2S = moles of Ag2S × molar mass of Ag2S
= 0.0082 mol × 247.8 g/mol (molar mass of Ag2S)
= 2.03 g
Therefore, 2.03 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.