Question

4. Copper metal reacts with nitric acid(HNO3) to produce aqueous copper (II)nitrate, nitrogen dioxide gas and liquid water. a. Write the balanced equation for the reaction b. If there are 0.500 moles of copper metal present How many moles of nitric acid are required for the reaction? How many moles of nitrogen dioxide gas are formed?

Answer 1

Answer:

a. $4HNO_3_(_a_q_)\hspace{0.1cm}+\hspace{0.1cm}Cu_(_s_)->Cu(NO_3)_2_(_a_q_)\hspace{0.1cm}+\hspace{0.1cm}2H_2O_(_l_)\hspace{0.1cm}+2NO_2_(_g_)$

b. $2molHNO_3\hspace{0.5cm}1molNO_2$

Explanation:

a. For the balanced equation we have to obtain the same amount of atoms in both sides:

$4HNO_3_(_a_q_)\hspace{0.1cm}+\hspace{0.1cm}Cu_(_s_)->Cu(NO_3)_2\hspace{0.1cm}+\hspace{0.1cm}2H_2O_(_l_)\hspace{0.1cm}+2NO_2_(_g_)$

b. If we want to calculate the moles of nitric acid $(HNO_3)$ we have to use the molar ratio in the balence reaction:

$0.5\hspace{0.1cm}mol\hspace{0.1cm}Cu_(_s_)\hspace{0.1cm}\frac{4\hspace{0.1cm}mol\hspace{0.1cm}HNO_3}{1\hspace{0.1cm}mol\hspace{0.1cm}Cu} = 2\hspace{0.1cm}mol\hspace{0.1cm}HNO_3_(_a_q_)$

For the moles of nitrogen dioxide $(NO_2)$ we have to follow the same logic:

$0.5\hspace{0.1cm}mol\hspace{0.1cm}Cu_(_s_)\hspace{0.1cm}\frac{2\hspace{0.1cm}mol\hspace{0.1cm}NO_2}{1\hspace{0.1cm}mol\hspace{0.1cm}Cu} = 1\hspace{0.1cm}mol\hspace{0.1cm}NO_2$

Answer 2

The balanced equation of the reaction is as follows:

Cu + 4HNO3 = Cu(NO3)2 + 2H2O+ 2NO2

1. 0.5 moles of Copper will react with 2 moles of HNO3
2. 0.5 moles of Copper will form 1 mole of NO2.

• According to this question, copper (Cu) reacts with HNO3 to produce copper nitrate, nitrogen dioxide and water. The balanced equation is as follows:

• Cu + 4HNO3 = Cu(NO3)2 + 2NO2 + 2H2O

1. 1 mole of Cu reacts with 4 moles of HNO3
2. Hence, 0.5 moles of Cu will react with (0.5 × 4) = 2 moles of HNO3.

1. Also, 1 mole of Cu forms 2 moles of NO2.
2. Hence, 0.5 moles of Cu will form (0.5 × 2) = 1 mole of NO2.

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