"2NO_2 \\rightarrow N_2O_4"
Convert g of "NO_2" to moles
"3.42 g (\\frac{1 mol}{46.01g})=0.0743 mol"
Covert moles of "NO_2" to moles of "N_2O_4"
"0.0743 mol NO_2(\\frac{1 mole N_2O_4}{2 moles NO_2})=0.0372 mol N_2O_4"
use Ideal Gas Law to find the volume of "N_2O_4"
"PV=nRT"
"V=\\frac{nRT}{P}=\\frac{0.0372\\times0.082 \\times (273+227)}{10.0}=0.153 L"
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