Respuesta :
Answer: 0.736 g
Explanation:
[tex]O_3+NO\rightarrow O_2+NO2[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of}O_3=\frac{0.781g}{48g/mol}=0.016moles[/tex]
[tex]\text{Number of moles of}NO=\frac{0.589g}{30g/mol}=0.019moles[/tex]
By Stoichiometry:
1 mole of ozone [tex]O_3[/tex] reacts with 1 mole of nitric oxide [tex]NO[/tex] to form 1 mole of nitrogen dioxide [tex]NO_2[/tex]
0.016 moles of ozone reacts with=[tex]\frac{1}{1}\times 0.016=0.016moles[/tex] of nitric oxide to form 0.016 mole of [tex]NO_2[/tex]
Thus ozone is the limiting reagent as it limits the formation of products and nitric oxide is the excess reagent as (0.019-0.016) g= 0.003 g remains as such.
Mass of [tex]NO_2=moles\times {\text{Molar mass}}=0.016\times 46=0.736g[/tex]
0.736 g of [tex]NO_2[/tex] will be produced.
