Cupper (II) sulfate reacts with hydrogen iodide to produce copper iodide, sulfuric acid, and iodine gas. If 10.4g of copper (II) sulfate is used, calculate the number of grams of HI needed and the number of grams of each product produced.
2CuSO4 + 4HI ==> 2CuI + 2H2SO4 + I2
Here is an example stoichiometry problem I've posted. Just follow the steps, one to determine HI, and separately calculate grams of each product formed.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To solve this problem, we need to balance the chemical equation first. The balanced equation for the reaction between copper (II) sulfate (CuSO4) and hydrogen iodide (HI) is:
CuSO4 + 4HI → CuI2 + H2SO4 + I2
The molar masses of the reactants and products involved in the reaction are:
- CuSO4: 159.61 g/mol
- HI: 127.91 g/mol
- CuI2: 367.01 g/mol
- H2SO4: 98.09 g/mol
- I2: 253.81 g/mol
Now, let's calculate the amount of HI required and the masses of each product.
1. Calculate the moles of CuSO4 using the given mass:
Moles of CuSO4 = Mass of CuSO4 / Molar mass of CuSO4
= 10.4 g / 159.61 g/mol
2. According to the balanced equation, 1 mole of CuSO4 reacts with 4 moles of HI. Therefore, the moles of HI required are:
Moles of HI = Moles of CuSO4 x (4 moles of HI / 1 mole of CuSO4)
3. Calculate the mass of HI:
Mass of HI = Moles of HI x Molar mass of HI
4. To find the masses of each product, we can use the stoichiometry of the balanced equation. 1 mole of CuSO4 produces 1 mole of CuI2, 1 mole of H2SO4, and 1 mole of I2:
Mass of CuI2 = Moles of CuSO4 x (1 mole of CuI2 / 1 mole of CuSO4) x Molar mass of CuI2
Mass of H2SO4 = Moles of CuSO4 x (1 mole of H2SO4 / 1 mole of CuSO4) x Molar mass of H2SO4
Mass of I2 = Moles of CuSO4 x (1 mole of I2 / 1 mole of CuSO4) x Molar mass of I2
Now you have the steps to calculate the number of grams of HI needed and the number of grams of each product produced. Just substitute the values in the equations using the molar masses provided.