How many grams of sulfur dioxide is released by the interaction of 6.4 g of copper
| April 11, 2021 | education
How many grams of sulfur dioxide is released by the interaction of 6.4 g of copper with an excess of concentrated sulfuric acid, if its yield is 85% of theoretically possible?
In order to solve this problem, we perform the following actions:
First, you need to draw up a reaction equation for this problem.
2Сu + 4 H2SO4 = 2SO2 + 2 CuSO4 + 2H2O
From the reaction equation we obtain
0.1mol – xmol
2mol – 2mol
we calculate the amount of copper and sulfur oxide and get the following:
n (Cu) = 6.4 / 64 = 0.1 mol
n (So2) = 0.1 × 2/2 = 0.1 mol
Next, we calculate the mass of sulfur oxide
mSo2 = 0.1 × 85% = 8.5g
mSo2 = 8.5 × 85 ÷ 100 = 7.22g
Answer: 7.22 grams of sulfur dioxide is released when copper interacts with concentrated sulfuric acid.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
