Question

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H20(1) a) If 5.58 g of copper(II) nitrate, Cu(NO3)2, is eventually obtained, how many grams of nitrogen monoxide, NO, would have formed?

Answer 1

Answer: 0.6 grams is the answer

Step-by-step explanation:

Answer 2

Answer: The mass of nitrogen monoxide formed is 0.6 grams.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For Copper (II) nitrate:

Given mass of copper (II) nitrate = 5.58 g

Molar mass of copper (II) nitrate = 187.56 g/mol

Putting values in above equation, we get:

`\text{Moles of copper (II) nitrate}=(5.58g)/(187.56g/mol)=0.03mol`

The given chemical equation follows:

`3Cu(s)+8HNO_3(aq.)\rightarrow 3Cu(NO_3)_2(aq.)+2NO(g)+4H_2O(l)`

By Stoichiometry of the reaction:

When 3 moles of copper (II) nitrate is formed, then 2 moles of nitrogen monoxide is formed.

So, when 0.03 moles of copper (II) nitrate is formed, then
`(2)/(3)* 0.03=0.02mol` of nitrogen monoxide is formed.

Now, calculating the mass of nitrogen monoxide by using equation 1, we get:

Molar mass of nitrogen monoxide = 30 g/mol

Moles of nitrogen monoxide = 0.02 moles

Putting values in equation 1, we get:

`0.02mol=\frac{\text{Mass of nitrogen monoxide}}{30g/mol}\\\\\text{Mass of nitrogen monoxide}=0.6g`

Hence, the mass of nitrogen monoxide formed is 0.6 grams.