How many grams of water are theoretically produced for the following reaction given we have 2.6 moles of hcl and 1.4 moles of ca(oh)2? reaction: 2hcl + ca(oh)2 → 2h2o + cacl2 a) 25.2 b) 50.4 c) 46.8 d) 103.6 e) not enough information

The amount of water is theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)₂ -  C) 46.8.

Given:

2HCl + Ca(OH)₂ -------------> CaCl₂ + 2H2O

in this reaction, 2 moles HCl reacts with 1 mole Ca(OH)₂

which means 2.6 moles of HCl reacts with 2.6 x = 1.3

but we have 1.4 moles Ca(OH)₂  it is the excess reagent

Solution:

HCl is a limiting reagent in this reaction as it allows producing the mole of water on the base of the number of moles it has,

=> 2 mole HCl - 2 moles of H₂O

Here, 2.6 moles of HCl would give 2.6 moles of H₂O only

=> convert moles to mass  

moles = mass / molar mass  

and,

mass =

mass = 2.6 x 18

= 46.8 g

Thus, the amount of water is theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)₂ -  C) 46.8.

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The answer to your question is: 6.8 g of water

Explanation:

Data

2.6 moles of HCl

1.4 moles of Ca(OH)2

                           2HCl     +     Ca(OH)2    →        2H2O    +      CaCl2

MW                   2(36.5)               74                       36 g               111 g

                          73g                

                            1 mol of HCl  36.5 g

                           2.6 mol              x

                              x = (2.6 x 36.5) / 1   = 94.9 g

                           1 mol of Ca(OH)2   74 g

                         1.4 mol                     x

                            x = (1.4 x 74) / 1  = 103.6 g

Grams of water

                        73 g of HCl   36g of H2O

                        94.9 g            x

                     x = (94.9 x 36) / 73 = 46.8 g of water