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87.58 L of C₂H₂

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.