3 Bu 24 Rt 4 Mm 6 Rk Ratio – 3:4:6 or 1:1.33:2 So… we would need 4.4 Rk 6 -> 24 3.5 -> x X=14

Step 1 – Use the ratio to find how much is needed of each reactant. 4 5 4 6 Step 1 – Use the ratio to find how much is needed of each reactant. 2.90 mol NH3 Step 2 – Compare with reactant needed to reactant available. 5 mol O2 3.625 mol O2 needed 4 mol NH3 3.75 mol O2 4 mol NH3 3 mol NH3 needed 5 mol O2 Must use limiting reactant! 2.90 mol NH3 4 mol NO 2.90 mol NO produced 4 mol NH3 3.75 mol O2 available - 3.625 mol O2 needed = 0.125 mol O2 excess

Na2SO3 limiting reactant Neither limits – enough of both reagents to fully react (under ideal conditions!) Na2SO3 limiting reactant 25.0 𝑔 𝑁𝑎2𝑆𝑂3 1 × 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 126.04 𝑔 𝑁𝑎2𝑆𝑂3 × 2 𝑚𝑜𝑙 𝐻𝐶𝑙 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 × 36.46 𝑔 𝐻𝐶𝑙 1 𝑚𝑜𝑙 𝐻𝐶𝑙 =14.46 𝑔 𝐻𝐶𝑙 𝑛𝑒𝑒𝑑𝑒𝑑 22.0 𝑔 𝐻𝐶𝑙 1 × 1 𝑚𝑜𝑙 𝐻𝐶𝑙 36.46 𝑔 𝐻𝐶𝑙 × 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 2 𝑚𝑜𝑙 𝐻𝐶𝑙 × 126.04 𝑔 𝑁𝑎2𝑆𝑂3 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 =38.0 𝑔 𝑁𝑎2𝑆𝑂3 𝑛𝑒𝑒𝑑𝑒𝑑

This is theoretical yield. WHY? Use limiting reactant! 25.0 𝑔 𝑁𝑎2𝑆𝑂3 1 × 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 126.04 𝑔 𝑁𝑎2𝑆𝑂3 × 1 𝑚𝑜𝑙 𝑆𝑂2 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 × 64.066 𝑔 𝑆𝑂2 1 𝑚𝑜𝑙 𝑆𝑂2 =12.7 𝑔 𝑆𝑂2 𝑐𝑎𝑛 𝑏𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 This is theoretical yield. WHY? % yield = 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑎𝑙𝑠𝑜 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑎𝑘𝑎 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑦𝑖𝑒𝑙𝑑 𝑥 100 % yield = 4.73𝑔 12.7𝑔 𝑥 100 =37.2%

When given mass: Step 1 – convert mass to moles Step 2 – Divide by smallest # to find ratio Step 3 – If cannot be rounded, multiply all by same number 2.3 𝑔 𝑁𝑎 1 × 1 𝑚𝑜𝑙 𝑁𝑎 23.0 𝑔 𝑁𝑎 =0.1𝑚𝑜𝑙𝑁𝑎 /0.05 0.8 𝑔 𝑂 1 × 1 𝑚𝑜𝑙 𝑂 16 𝑔 𝑂 =0.05𝑚𝑜𝑙𝑂 /0.05 =1 =2 Empirical formula = Na2O

When given percentages, assume 100g, then take normal steps Empirical Molecular When given percentages, assume 100g, then take normal steps Ratio = 𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒆𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 85.7 𝑔 𝐶 1 × 1 𝑚𝑜𝑙 𝐶 12.0 𝑔 𝐶 =7.08𝑚𝑜𝑙𝐶 /7.08 Ratio = 28 (12+2) =2 =1 Empirical formula x 2 = C2H4 14.3 𝑔 𝐻 1 × 1 𝑚𝑜𝑙 𝐻 1 𝑔 𝐻 =14.3𝑚𝑜𝑙𝐻 /7.08 ≈𝟐 Empirical formula = CH2