Respuesta :
Answer: The mass of nitrogen monoxide formed is 0.6 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(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:
[tex]\text{Moles of copper (II) nitrate}=\frac{5.58g}{187.56g/mol}=0.03mol[/tex]
The given chemical equation follows:
[tex]3Cu(s)+8HNO_3(aq.)\rightarrow 3Cu(NO_3)_2(aq.)+2NO(g)+4H_2O(l)[/tex]
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 [tex]\frac{2}{3}\times 0.03=0.02mol[/tex] 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:
[tex]0.02mol=\frac{\text{Mass of nitrogen monoxide}}{30g/mol}\\\\\text{Mass of nitrogen monoxide}=0.6g[/tex]
Hence, the mass of nitrogen monoxide formed is 0.6 grams.
