16 Best Examples on Stoichiometry table
The best way of understanding stoichiometry table is simply to use stoichiometry worksheet.
I’m going to use stoichiometry table to show some calculations on different aspects of stoichiometry. by illustrating with examples.
In the stoichiometry table, I’ll solve questions on these categories just to basically help us understand stoichiometry concept better..
- Mole to Mole stoichiometry
- Mole to mass stoichiometry
- Mole to volume stoichiometry
- Mass to mass stoichiometry
- Mass to volume stoichiometry
- Volume to volume stoichiometry
Calculations using Stoichiometry table
| MOLE TO MOLE STOICHIOMETRY | ||
| Questions | Answer | |
| 1 | What is the mole of Zinc produced by reacting 3.9 moles of magnesium? Mg + ZnCl2===MgCl2 + Zn | 1 mole of Mg === 1 mole of zinc 3.9 moles of Mg=== x X = 3.9 x 1 = 3.9 moles of zinc |
| 2 | What’s the mole of methane that will produce 1.5 moles of carbon iv oxide according to this equation? CH4 + 2O2 === CO2 + 2H2O? | 1 mole of CH4 == 1 mole of CO2 X moles of CH4== 1.2 mole of CO2 X= 1 mole x 1.5 mole /1mole =1.5 1.5 moles of methane |
| 3 | How many moles of Propane would be needed to produce four moles of carbon iv oxide according to this C3H8 + 5O2===3CO2 + 4H2O | C3H8 + 5O2===3CO2 + 4H2O 1 mole of C3H8 ==3 moles of CO2 X moles of C3H8==4 moles of CO2 X= 1 mole X 4 moles /3 moles × = 1.667 mole of carbon iv oxide evolved X= 1.33 moles of carbon iv oxide |
| MOLE TO MASS SSTOICHIOMETRY | ||
| 4 | What is the mass of Calcium chloride produced when 3.5 moles of Calcium reacts with iron II chloride according to this equation Ca + ZnCl2===CaCl2 + Zn | I mole of Ca ==1 mole of CaCl2 3.5 moles of Ca == x X = 3.5 x 1/1 = 3.5 moles of CaCl2 Then we convert the mole to mass Since I mole of CaCl2== 111g of CaCl2 Mass of CaCl2 = (40 +71=95g) Therefore 3.5 mole of CaCl2= x X= 3.5 x 111=388.5g of Calcium Chloride |
| 5 | What mass of carbon monoxide is would be evolved by the incomplete combustion of 4.2 moles of carbon according to this equation 2C + O2===2CO? | 2C + O2====2CO 2 moles of C == 2 moles of CO 4.2 moles of C == x X = 4.2 x 2/2 = 4.2 moles of CO Then we convert the mole to mass Since I mole of CO===== 28g of CO (12 +16=28g) Therefore 4.2 mole of CO=== x X= 4.2 x 28=117.6g Carbon monoxide |
| 6 | What is the mass of potassium hydroxide produced by reacting 2.5 moles of potassium with cold water? 2K + 2H2O ==2KOH + H2 | 2K + 2H2O ==2KOH + H2 2 moles of K === 2 moles of KOH 2.5 moles of K ===== x X = 2.5 x 2/2 = 2.5 moles of KOH Then we convert the mole to mass Since I mole of KOH=== 56g of CO (39 +16+ 1)=56g Therefore 2.5 moles of KOH===x X= 2.5 x 56=140 g of potassium hydroxide |
| MOLE TO VOLUME STOICHIOMETRY | ||
| 7 | What volume of chlorine gas would be evolved by reacting 3.3 moles of fluorine with sodium chloride F2 + 2NaCl ==2NaF + Cl2 | F2 + 2NaCl ==2NaF + Cl2 1 mole of F2 === 1 mole of Cl2 3.3 moles of F2 === x X = 3.3 x 1/1 = 3.3 moles of Cl2 Then we convert the mole to mass Since I mole of Cl2=== 22.4 dm 3.3 moles of Cl2=== x X = 73.92 dm3 |
| 8 | What volume of oxygen would react with 0.95 mole of sodium to form sodium oxide 4Na + O2===2Na2O | 4Na + O2===2Na2O 4 moles of Na === 1 mole of O2 0.95 moles of Na === x X = 0.95 x 1/4 = 0.2375 mole of O2 Then we convert the mole to mass Since I mole of O2=== 22.4 dm 0.2375 mole of O2=== x X = 0.2375 X 22.4 = 5.249 dm3 |
| 9 | What volume of ethyne is evolved by reacting with 2.5 moles of calcium carbide with water CaC2 + 2H2O ==Ca(OH)2+ C2H2 | CaC2 + 2H2O ==Ca(OH)2+ C2H2 1 mole of CaC2 === 1 mole of C2H2 2.5 moles of CaC2=== x X = 0.95 x 1/4 = 0.2375 mole of O2 Then we convert the mole to mass Since I mole of O2=== 22.4 dm 0.2375 mole of O2=== x X = 0.2375 X 22.4 = 5.249 dm3 |
| MASS TO MASS STOICHIOMETRY | ||
| 10 | What mass of carbon monoxide is evolved by 20g of carbon burning in oxygen 2C + O2 ===2CO | 2C + O2 ===2CO 2 moles of C === 2 moles of CO Converting to mass 2 x 12g === 2 x 28g 24 grams of C ===56 grams of CO 20 grams of C === x X = 24 x 56 /20 X = 46.67 grams of CO |
| 11 | If 30 grams of nitrogen reacted with hydrogen. Calculate the mass of ammonia that would be evolved N2 + 3H2 ==2NH3 | N2 + 3H2 ==2NH3 1 mole of N2 === 2 moles of NH3 Converting to mass 28 grams of N2=== 34 grams of NH3 30 grams of N2 === x X = 30 x 34 / 28 X = 36.4 grams of Ammonia |
| 12 | 58 grams of calcium carbonate decomposed into calcium oxide and carbon iv oxide. What mass of Carbon iv oxide is evolved? CaCO3 ==CaO + CO2 | CaCO3 ==CaO + CO2 1 mole of CaCO3 === 1 moles of CO2 Converting to mass 100 grams of CaCO3=== 44 grams of CO2 58 grams of CaCO3=== x X = 58 x 44 / 100 X = 25.52 grams of Carbon IV oxide |
| MASS TO VOLUME STOICHIOMETRY | ||
| 13 | If 29 grams of Potassium hydrogen trioxocarbonate IV decomposed. Calculate the volume of oxygen evolved 2KHCO3 ===K2CO3 + H2O +O2 | 2KHCO3 ===K2CO3 + H2O +O2 2 moles of KHCO3 === 1 mole of O2 2 X100 grams of KHCO3=== 32 grams of O2 200 grams of KHCO3=== 32 grams of O2 29 grams of KHCO3=== x X = 29 x 32 / 200 X = 4.64 grams of Oxygen |
| 14 | 18.8 grams of Nitrogen reacted with Hydrogen N2 + 3H2 ===2NH3 Calculate the volume of ammonia evolved. | N2 + 3H2 ===2NH3 1 mole of N2 === 2 mole of NH3 28 grams of N2=== 44.8 dm3 of NH3 18.8 grams of N2=== x X = 30.08 dm3 of Ammonia |
| VOLUME TO VOLUME STOICHIOMETRY | ||
| 15 | What volume of sulphur vi oxide is evolved by reacting 12.6 dm3 of sulphur iv oxide. 2SO2 + O2 ==2SO3 | 2SO2 + O2 ==2SO3 2 moles of SO2=== 2 moles of SO3 Converting moles to volume 44.8 dm3 of SO2=== 44.8 dm3 of SO3 12.6 dm3 of SO2=== x X = 12.6 x 44.8 /44.8 X = 12.6 dm3 |
| 16 | What volume of carbon iv oxide is evolved by burning 300cm3of methane? CH4 + 2O2===CO2 +2H2O | CH4 + 2O2===CO2 +2H2O 1 moles of CH4=== 1 mole of CO2 Converting moles to volume 22.4 dm3 of CH4=== 22.4 dm3 of CO2 300 cm3 means 3 dm3 of CH4 3 dm3 of CH4 === x X = 3 x 22.4 /22.4 X = 3 dm3 |
So, stoichiometry table worksheet above illustrates the different steps and practice questions necessary in solving stoichiometry problems.
