Определить объем ацетилена, кот. образуется при взаимодействии 6,4 г. карбида кальция и 1,8 г. воды, если выход 90%.
To calculate the volume of acetylene produced when 6.4 grams of calcium carbide and 1.8 grams of water react with a yield of 90%, you can follow these steps:1. Write a balanced chemical equation for the reaction between calcium carbide (CaC2) and water (H2O):
CaC2 + 2H2O → C2H2 + Ca(OH)2
2. Determine the molar mass of calcium carbide and water:
CaC2: 1 mol CaC2 = 64 grams
H2O: 2 mol H2O = 36 grams
3. Convert the given masses of calcium carbide and water into moles:
6.4 grams CaC2 x (1 mol CaC2/64 grams CaC2) = 0.1 mol CaC2
1.8 grams H2O x (1 mol H2O/18 grams H2O) = 0.1 mol H2O
4. Use the mole ratios from the balanced chemical equation to determine the theoretical yield of acetylene:
Mole ratio of CaC2 to C2H2: 1:1
0.1 mol CaC2 → 0.1 mol C2H2
5. Calculate the volume of acetylene using the ideal gas law:
V = nRT/P
n = moles of acetylene = 0.1 mol
R = ideal gas constant = 0.0821 L·atm/mol·K
T = temperature = 273 K
P = pressure = 1 atm
Substituting the values:
V = (0.1 mol)(0.0821 L·atm/mol·K)(273 K)/(1 atm) = 2.255 L
Therefore, the volume of acetylene produced is 2.255 liters.
It's worth mentioning that this is the theoretical yield and the actual yield may be slightly different due to experimental errors.
Now, if you need to convert the answer to a different unit, you can do so by using a conversion factor. For example, if you want the answer in milliliters (mL), you can use the conversion factor 1 L = 1000 mL, giving an answer of 2255 mL.
In conclusion, by following these steps and using the ideal gas law, you can determine the volume of acetylene produced when 6.4 grams of calcium carbide and 1.8 grams of water react with a yield of 90%.
