Copper reacts with sulfuric acid according to the following equation: 2H3SO4 + Cu-> CuSO4 + 2H2O + SO2 How many grams of sulfur dioxide are created by this reaction if 14.2 g of copper reacts with 18 g of sulfuric acid?
You have a limiting regent problem here. Basically, you work two stoichiometry problems and take the one producing the smaller amount of product as the valid one.
1. Convert 14.2 g Cu to moles. moles = grams/molar mass
2. Convert 18 g H2SO4 to moles using the same process.
3. Using the coefficients in the balanced equation, convert moles Cu to moles of the product.
4. Same procedure, convert moles H2SO4 5o moles of the product.
5. It is likely that the answer from 3 and 4 will not agree. The correct answer, in limiting reagent problems. is ALWAYS the smaller one.
6. Convert mole SO2 from step 5 to grams. g = moles x molar mass.
Post your work if you get stuck.
I don't understand step 3?
To determine the amount of sulfur dioxide (SO2) produced when 14.2 g of copper reacts with 18 g of sulfuric acid (H3SO4), we need to use the given equation:
2H3SO4 + Cu -> CuSO4 + 2H2O + SO2
First, calculate the molar mass of copper (Cu):
- The molar mass of copper (Cu) is 63.55 g/mol.
Next, calculate the molar mass of sulfuric acid (H3SO4):
- Hydrogen (H) has a molar mass of 1.01 g/mol
- Sulfur (S) has a molar mass of 32.07 g/mol
- Oxygen (O) has a molar mass of 16.00 g/mol
- The molar mass of sulfuric acid (H3SO4) is (3*1.01) + 32.07 + (4*16.00) = 98.09 g/mol.
Then, calculate the number of moles of copper (Cu) by dividing its mass by its molar mass:
- Moles of Cu = 14.2 g / 63.55 g/mol = 0.2238 mol.
Calculate the number of moles of sulfuric acid (H3SO4) by dividing its mass by its molar mass:
- Moles of H3SO4 = 18 g / 98.09 g/mol = 0.1835 mol.
According to the balanced equation, the ratio of moles of copper (Cu) to moles of sulfur dioxide (SO2) is 1:1. This means that for every 1 mole of copper reacting, 1 mole of sulfur dioxide (SO2) is produced.
Finally, calculate the mass of sulfur dioxide (SO2) produced by multiplying the number of moles of SO2 by its molar mass:
- Moles of SO2 = 0.1835 mol.
- Mass of SO2 = Moles of SO2 * Molar mass of SO2 = 0.1835 mol * 64.06 g/mol = 11.8 g.
Therefore, 11.8 grams of sulfur dioxide (SO2) are created by this reaction.
To find out how many grams of sulfur dioxide (SO2) are produced by the reaction, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
To identify the limiting reactant, we can use stoichiometry and compare the amounts of copper and sulfuric acid in the given reaction. The balanced equation tells us that 1 mole of copper reacts with 2 moles of sulfuric acid to produce 1 mole of sulfur dioxide.
1. Calculate the moles of copper (Cu) and sulfuric acid (H3SO4):
Moles of Cu = Mass of Cu / Molar mass of Cu
Moles of Cu = 14.2 g / 63.55 g/mol (molar mass of Cu) = 0.2238 mol
Moles of H3SO4 = Mass of H3SO4 / Molar mass of H3SO4
Moles of H3SO4 = 18 g / (1 g/mol + 3(16 g/mol) + 4(1 g/mol)) (molar mass of H3SO4) = 0.1481 mol
2. Use the stoichiometry ratio to compare the moles of copper and sulfuric acid. The stoichiometry tells us that 1 mole of copper reacts with 2 moles of sulfuric acid to produce 1 mole of sulfur dioxide.
Moles of Cu / Stoichiometry ratio = 0.2238 mol / 1 = 0.2238 mol
Moles of H3SO4 / Stoichiometry ratio = 0.1481 mol / 2 = 0.0741 mol
Since there is less moles of sulfuric acid (0.0741 mol) compared to the moles of copper (0.2238 mol), sulfuric acid is the limiting reactant.
3. Determine the amount of sulfur dioxide produced using the moles of limiting reactant (sulfuric acid) and the stoichiometry ratio:
Moles of SO2 = Moles of H3SO4 / Stoichiometry ratio
Moles of SO2 = 0.0741 mol / 2 = 0.0370 mol
4. Finally, calculate the mass of sulfur dioxide using the moles and the molar mass of SO2:
Mass of SO2 = Moles of SO2 * Molar mass of SO2
Mass of SO2 = 0.0370 mol * (32.07 g/mol) (molar mass of SO2)
Mass of SO2 = 1.19 g
Therefore, 1.19 grams of sulfur dioxide are produced by the reaction.