1. 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

2. 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 mol O2 needed =
0.125 mol O2 excess

3. 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 𝑁𝑎2𝑆𝑂 𝑔 𝑁𝑎2𝑆𝑂3 × 2 𝑚𝑜𝑙 𝐻𝐶𝑙 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 × 𝑔 𝐻𝐶𝑙 1 𝑚𝑜𝑙 𝐻𝐶𝑙 =14.46 𝑔 𝐻𝐶𝑙 𝑛𝑒𝑒𝑑𝑒𝑑
22.0 𝑔 𝐻𝐶𝑙 1 × 1 𝑚𝑜𝑙 𝐻𝐶𝑙 𝑔 𝐻𝐶𝑙 × 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 2 𝑚𝑜𝑙 𝐻𝐶𝑙 × 𝑔 𝑁𝑎2𝑆𝑂3 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 =38.0 𝑔 𝑁𝑎2𝑆𝑂3 𝑛𝑒𝑒𝑑𝑒𝑑

4. This is theoretical yield. WHY?
Use limiting reactant!
25.0 𝑔 𝑁𝑎2𝑆𝑂3 1 × 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂 𝑔 𝑁𝑎2𝑆𝑂3 × 1 𝑚𝑜𝑙 𝑆𝑂2 1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂3 × 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%

6. 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

7. 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