## Introductory Chemistry (5th Edition)

a) $2.6 moles NO_{2}$ b) $11.6 moles NO_{2}$ c) $8900 moles NO_{2}$ d) $2.012\times10^{-3} moles NO_{2}$
We can solve this by looking at the balanced chemical equation and using the mole ratios: $2N_{2}O_{5} --> 4NO_{2} + O_{2}$ a) $1.3 moles N_{2}O_{5} \times\frac{4 moles NO_{2}}{2 moles N_{2}O_{5}} = 2.6 moles NO_{2}$ b) $5.8 moles N_{2}O_{5} \times\frac{4 moles NO_{2}}{2 moles N_{2}O_{5}} = 11.6 moles NO_{2}$ c) $4.45\times10^{3} moles N_{2}O_{5} \times\frac{4 moles NO_{2}}{2 moles N_{2}O_{5}} = 8900 moles NO_{2}$ d) $1.006\times10^{-3} moles N_{2}O_{5} \times\frac{4 moles NO_{2}}{2 moles N_{2}O_{5}} = 2.012\times10^{-3} moles NO_{2}$