At 2.85 atm and 484 K, what volume of NO2 gas forms from 48.18 g of O2? 4NH3(g)+7O2(g)⟶4NO2(g)+6H2O(l)

Answer 1

1) Convert 48.18 g of O₂ into moles of O₂

First find the molecular weight of O₂

2 (because 2 O)*16 g=32g/mol of O₂

48.18g*(1mol/32g)=1.51 mol O₂

2) Use the answer from part 1 to convert mol of O₂ to mol of NO₂

1.51 mol O2 *(4 mol NO₂/ 7 mol O₂)=0.860 mol NO₂

3) Use PV=nRT to solve for volume

Rearrange:

V=nRT/P

V=0.860 mol* 0.082 L*atm/(K*mol)*484K/2.85atm=11.98 L

*R is the universal gas constant there are different values for this constant dependent on the units assigned to it. In this case I used 0.082 and calculated the answer in liters.