Question

Calcium carbide reacts with water to produce acetylene gas. A chemist in a quality control lab is testing the purity of calcium carbide. She measures 9.37 g of grey solid from a tin labelled “calcium carbide”, and reacts it with water to collect 935 mL of acetylene gas at 99.5 kPa and 17.0 °C. What percentage of her original grey solid was calcium carbide?

CaC2(s) + 2 H2O(l) → C2H2(g)

Answer 1

CaC₂+2H₂O=C₂H₂+Ca(OH)₂

1.The amount of C₂H₂ can be found by the following Mendeleev-Clapeyron equation:

n=P*V/R*T where R is equal to 8.31

n=99.5*0.935/8.31*290 =0.0386mol

Comment: 290 is the temperature in Kelvin. To find this, 273 degrees Celsius is added

2.According to the above reaction, 0.0386 mol of C₂H₂ is formed from as many moles of CaC₂. Hence 0.0386mol(2.4704gr) CaC₂ was reacted

3.determine the percentage of CaC₂ unit in the starting material. To do this, the mass of CaC₂ is divided by the mass of the total substance and multiplied by 100%

2.47/9.37=0.26365*100%=26.365%

Answer: 26.365%