Answer:
87.58 L of C₂H₂
Step-by-step explanation:
We'll begin by calculating the number of mole in 250 g of CaC₂.
This can be obtained as follow:
Mass of CaC₂ = 250 g
Molar mass of CaC₂ = 40 + (12×2)
= 40 + 24
= 64 g/mol
Mole of CaC₂ =?
Mole = mass /Molar mass
Mole of CaC₂ = 250 / 64
Mole of CaC₂ = 3.91 moles
Next, the balanced equation for the reaction. This is given below:
CaC₂ + 2H₂O —> C₂H₂ + Ca(OH)₂
From the balanced equation above,
1 mole of CaC₂ reacted to produce 1 mole of C₂H₂.
Next, we shall determine the number of mole of C₂H₂ produced by the reaction of 250 g (i.e 3.91 moles) of CaC₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaC₂ reacted to produce 1 mole of C₂H₂.
Therefore, 3.91 moles of CaC₂ will also react to produce 3.91 moles of C₂H₂.
Finally, we shall determine the volume of C₂H₂ produced from the reaction. This can be obtained as follow:
Recall:
1 mole of any gas occupy 22.4 L at STP.
1 mole of C₂H₂ occupied 22.4 L at STP.
Therefore, 3.91 moles of C₂H₂ will occupy = 3.91 × 22.4 = 87.58 L
Thus, 87.58 L of C₂H₂ is produced from the reaction.