Calcium carbide ( CaC2) reacts with water to produce acetylene (C2H2) as shown in the unbalanced reaction below: CaC2(s)+H2O(g)->Ca(OH)2(s)+C2H2(g). How many grams of H2O are required to completely react with 82.4g of CaC2

Answers

Answer 1
Answer:
46.3g H2O

Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)

then use factor label method to solve

82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
Answer 2

The given reaction can be balanced as follows: CaC2 + 2H2O → Ca (OH)2 + C2H2. Grams of water required to completely react with 82.4 g of CaC2 is 46.28 g.

What is a balanced reaction?

In balanced reaction, the number of each and every element on both side is balanced. In the given reaction to balance the number of H and O, add 2 to H₂O so that, all the species will be equal in number on both side.

Now, as per the balanced reaction, 2 moles of water is required to reacts with 1 mole of CaC2.

Mass of one mole of water = 18 g/mol.

                                   2 mole = 36 g

Mass of one mole of CaC2 = 64.09 g/mol

Thus 36 g of water is required to react with 64.09 grams of calcium carbide. Mass of water required to react with 82.4 g of calcium carbide is calculated as follows:

Mass of water = (36 g×82.4 g)/64.09 g = 46.28 g

Hence, 46.28 grams of water is required to completely react with 82.4 g of calcium carbide.

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