hydrochloric acid is added to 50 g of iron (III) sulfide. what weight of hydrogen sulfide is produced? FeS+2HNO3+Mg(NO3)2+H2
You have a lousy equation. All pluses. No arrow. No way to tell which are the reactants and which the products. I'll leave that for you to correct but here is how you work the problem (in general terms).
1. Write and balance a good equation.
2. Convert 50 g Fe to moles. moles = grams/molar mass.'
3. Using the coefficients in the balanced equation, convert moles Fe to moles H2S.
4. Now convert moles H2S to grams. g = moles x molar mass.
To determine the weight of hydrogen sulfide produced in the reaction between hydrochloric acid (HCl) and iron (III) sulfide (FeS), we need to balance the chemical equation first. However, in your question, there are additional compounds mentioned (2HNO3, Mg(NO3)2, and H2), which seem unrelated to the reaction between HCl and FeS.
To solve the specific problem of iron (III) sulfide reacting with hydrochloric acid, we can write the balanced chemical equation:
FeS + 2HCl → FeCl2 + H2S
From the balanced equation, we can see that 1 mole of FeS reacts with 2 moles of HCl to produce 1 mole of H2S. To determine the weight of H2S produced, we need to use the molar mass of FeS and the molar ratio between FeS and H2S.
The molar mass of FeS (iron (III) sulfide) can be calculated as follows:
1 atom of iron (Fe) has a molar mass of approximately 55.85 g/mol, while 1 atom of sulfur (S) has a molar mass of approximately 32.06 g/mol.
So, the molar mass of FeS is:
Molar mass of FeS = (1 × 55.85 g/mol) + (1 × 32.06 g/mol)
Molar mass of FeS = 87.91 g/mol
Now we can set up the conversion using the molar ratio between FeS and H2S:
1 mole of FeS = 1 mole of H2S
1 mole of FeS = 87.91 g
To find the weight of H2S produced, we need to know the amount of FeS used in the reaction. You mentioned that 50 g of FeS was used. Thus, we can proceed with the calculation:
50 g FeS × (1 mole FeS / 87.91 g FeS) × (1 mole H2S / 1 mole FeS) × (34.08 g H2S / 1 mole H2S)
Calculating this expression will give you the weight of hydrogen sulfide (H2S) produced in grams.
To determine the weight of hydrogen sulfide produced when hydrochloric acid is added to iron (III) sulfide, we need to balance the chemical equation and use stoichiometry.
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and iron (III) sulfide (FeS) can be written as follows:
FeS + 2HCl → FeCl2 + H2S
From the balanced equation, we can see that 1 mole of iron (III) sulfide reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen sulfide.
To find the weight of hydrogen sulfide produced, we need to convert the given mass of iron (III) sulfide (FeS) to moles and then use the mole ratio from the balanced equation.
First, we need to determine the molar mass of iron (III) sulfide (FeS):
- The atomic mass of iron (Fe) is approximately 55.85 g/mol.
- The atomic mass of sulfur (S) is approximately 32.06 g/mol.
Molar mass of FeS = 55.85 g/mol + 32.06 g/mol = 87.91 g/mol
Next, let's calculate the number of moles of FeS:
- Given mass of FeS = 50 g
- Moles of FeS = mass of FeS / molar mass of FeS
Moles of FeS = 50 g / 87.91 g/mol ≈ 0.57 mol
Now, using the mole ratio from the balanced equation, we can determine the number of moles of hydrogen sulfide (H2S) produced:
- Moles of H2S = Moles of FeS × (1 mol H2S / 1 mol FeS)
- Moles of H2S = 0.57 mol × (1 mol H2S / 1 mol FeS) ≈ 0.57 mol
Finally, to find the weight of hydrogen sulfide produced, we multiply the number of moles of H2S by its molar mass:
- The molar mass of hydrogen sulfide (H2S) is approximately 34.08 g/mol.
Weight of H2S = Moles of H2S × Molar mass of H2S
Weight of H2S = 0.57 mol × 34.08 g/mol ≈ 19.45 g
Therefore, approximately 19.45 grams of hydrogen sulfide will be produced when hydrochloric acid is added to 50 grams of iron (III) sulfide.