When copper (ii) chloride reacts with sodium nitrate, copper (ii) nitrate and sodium chloride are formed. a) write the balanced equation for the reaction given above: b) if 45 grams of copper (ii) chloride react with 25 grams of sodium nitrate, how much c) what is the limiting reagent for the reaction? d) how many grams of copper(ii) nitrate is formed? e) how much of the excess reagent is left over in this reaction? f) if 11.3 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction?

The answer to your question is:

Explanation:

a)    CuCl₂   +  2NaNO₃    ⇒     Cu(NO₃)₂   + 2NaCl

b)

45 g CuCl₂

25 g NaNO₃                 Is not complete the question

c) MW CuCl₂ = 64 + 71 = 135 g

   MW NaNO₃ = 23 + 14 + 48 = 85 x 2 = 170 g

theoretical Ratio      170 / 135 = 1.26

experimental ratio   25/ 45 = 0.56

Limiting reactant = NaNO₃

d)                    170 g of NaNO₃  126 g of Cu(NO₃)₂

                    25 g                    x

                  x = (25 x 126)/170

                  x = 18.52 g of Cu(NO₃)₂

e)                170 g of NaNO₃    135 g of CuCl₂

                    25 g of NaNO₃    x

                   x = 19.85 g of CuCl₂

Excess = 45 -  19.85 = 25.15 g of CuCl₂

f) 11.3 g of NaCl

Theoretical

                    170 g of NaNO₃  117 g of NaCl

                      25 g                   x

                     x = (25 x 117) / 170

                     x = 17.2 g of NaCl

yield = 11.3/17.2 x 100

yield = 65.7 %

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol