If 4.50 moles of HNO3 are used, how many moles of copper (II) nitrate are produced?

2.25 moles of Cu(NO₃)₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Cu + 2HNO₃ —> Cu(NO₃)₂ + H₂

From the balanced equation above,

2 moles of HNO₃ reacted to produce 1 mole of Cu(NO₃)₂.

Finally, we shall determine the number of mole of Cu(NO₃)₂ produced by the reaction of 4.5 moles of HNO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of HNO₃ reacted to produce 1 mole of Cu(NO₃)₂.

Therefore, 4.5 moles of HNO₃ will react to produce = 2.25 moles of Cu(NO₃)₂.

Thus, 2.25 moles of Cu(NO₃)₂ were obtained from the reaction.

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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