If 18.5 grams of CuCl2 react with 22.8 grams of NaNO3, what mass of NaCl can be formed? CuCl2 + NaNO3 → Cu(NO3)2 + Naci
A. 12.4 g Naci
B. 15.7 g Naci
C. 16.2 g Naci
D. 28.4 g Naci

B =  15.7 g

Explanation:

Given data;

Mass of CuCl₂ = 18.5 g

mass of NaNO₃ = 22.8 g

Mass of NaCl produced = ?

Solution:

Chemical equation;

CuCl₂  + 2NaNO₃    →    Cu(NO₃)₂   +   2NaCl

Number of moles of CuCl₂:

Number of moles = mass/ molar mass

Number of moles = 18.5 g / 134.45 g/mol

Number of moles = 0.14 mol

Number of moles of NaNO₃:

Number of moles = mass/ molar mass

Number of moles = 22.8 g / 85 g/mol

Number of moles = 0.27 mol

Now we will compare the moles of NaCl with both of reactant.

                 CuCl₂             :            NaCl

                   1                    :              2

                  0.14                :            2/1×0.14 = 0.54

                  NaNO₃          :            NaCl

                    2                    :              2

                  0.27                :             0.27

less number of moles of NaCl are produced by  NaNO₃ it will act as limiting reactant.

Mass of NaCl produced:

Mass = number of moles × molar mass

Mass = 0.27 mol ×  58.4 g/mol

Mass =  15.7 g

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.