Solutior G potassium hydroxideGoogle x nttakeCovalentActivit

Solutior ×(G, potassium hydroxide-Google x nt/takeCovalentActivity.do?locatorassignment-take&takeAssignmentSessionLocator-assignment-take; a Use the References to access important values if needed for this question. Nitrogen dioxide is produced by combustion in an auto moles of nitrogen monoxide are mixed with 0.464 moles of oxygen gas. mobile engine. For the following reaction, 0.232 nitrogen monoxide(g) + oxygen(g) nitrogen dioxide(g) What is the formula for the limiting reagent? What is the maximum amount of nitrogen dioxide that can be produced? moles Retry Entire Group 4 more group attempts remaining

Solution

the balanced reaction is given by

2 NO + O2 --> 2 NO2

now

moles of NO required = 0.464 moles 02 x 2 mol NO / 1 mol O2

moles of NO required = 0.928 moles NO

but

only 0.232 moles of NO are present

so

NO is the limiting reagent

2)

now

2 NO + O2 ---> 2 NO2

moles of NO2 that can be produced = 0.232 moles NO x 2 mol NO2 / 2 mol NO

moles of NO2 that can be produced = 0.232 moles NO2

so

0.232 moles of nitrogen dioxide can be produced