Question

Use the balanced equation to solve the problem.N2 + 3H22NH323.0g NH3 are made.How many liters of H₂ gas reacted at Stp? L

Answer 1

`N_(2(g))+3H_(2(g))\rightarrow2NH_(3(g))`

By using the ideal gas law to get volume we have"

`V=(nRT)/(P)`

Where v is volume, T is temperatute, n is number of moles, R is the molar gas constant and P is pressure. At STP P= 101,325 Pa, T= 273.15 K and R= 8.314 J/mol K

`\begin{gathered} (RT)/(P)=0.022414cm^3mol^(-1) \\ \\ V=0.0022414n \end{gathered}`

We must first convert mass to moles:

`\begin{gathered} mole=\frac{mass}{molecular\text{ }mass} \\ mole=\frac{23.0g}{17.0g\text{ }mol^(-1)} \\ \\ mole=1.35 \end{gathered}`
`To\text{ }determine\text{ }the\text{ }moles\text{ }of\text{ }H2\text{ }gas\text{ }reacted\text{ }we:(2)/(3)*1.35=0.87\text{ }mol`

By substituting this value into the ideal gas law we have:

`\begin{gathered} V=0.0022414cm^3mol^(-1)*0.87mol \\ V=0.0019502cm^3 \\ \\ V=1.9502*10^(-6)L \end{gathered}`

1.9502e-6L of H2 gas reacted at STP