What volume at nitrogen would be required to react with 0.100g of hydrogen to produce ammomia?
You use the coefficients just as you did in dimensional analysis.
N2 + 3H2 ==> 2NH3.
What volume of NH3 will be formed with 1 g hydrogen. 1 mole H2 has a mass of 2 grams and it has a volume of 22.4L. Therefore, 1 g hydrogen will have a volume of 11.2 L (22.4L/mol) x (1 g) x (1mol H2/2g H2) = 11.2 L).
So 11.2 L H2 x (2 moles NH3/3 moles H2) = 7.467 L which would round to 7.47L NH3 formed.