# 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 + ZnCl _{2}===MgCl_{2} + 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? CH_{4 }+ 2O_{2} === CO_{2} + 2H_{2}O? | 1 mole of CH _{4} == 1 mole of CO_{2}X moles of CH _{4}== 1.2 mole of CO_{2}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 C_{3}H_{8} + 5O_{2}===3CO_{2} + 4H_{2}O | C_{3}H_{8} + 5O_{2}===3CO_{2} + 4H_{2}O 1 mole of C_{3}H_{8} ==3 moles of CO_{2} X moles of C _{3}H_{8}==4 moles of CO_{2} 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 + ZnCl_{2}===CaCl_{2} + Zn | I mole of Ca ==1 mole of CaCl_{2} 3.5 moles of Ca == x X = 3.5 x 1/1 = 3.5 moles of CaCl _{2} Then we convert the mole to mass Since I mole of CaCl _{2}== 111g of CaCl_{2} Mass of CaCl _{2} = (40 +71=95g) Therefore 3.5 mole of CaCl _{2}= 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 + O_{2}===2CO? | 2C + O_{2}====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 + 2H _{2}O ==2KOH + H_{2} | 2K + 2H _{2}O ==2KOH + H_{2} 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 F_{2 }+ 2NaCl ==2NaF + Cl_{2} | F _{2 }+ 2NaCl ==2NaF + Cl_{2}1 mole of F _{2} === 1 mole of Cl_{2} 3.3 moles of F _{2} === x X = 3.3 x 1/1 = 3.3 moles of Cl _{2} Then we convert the mole to mass Since I mole of Cl _{2}=== 22.4 dm 3.3 moles of Cl _{2}=== x X = 73.92 dm ^{3} |

8 | What volume of oxygen would react with 0.95 mole of sodium to form sodium oxide 4Na + O _{2}===2Na_{2}O | 4Na + O _{2}===2Na_{2}O4 moles of Na === 1 mole of O _{2} 0.95 moles of Na === x X = 0.95 x 1/4 = 0.2375 mole of O _{2} Then we convert the mole to mass Since I mole of O _{2}=== 22.4 dm 0.2375 mole of O _{2}=== x X = 0.2375 X 22.4 = 5.249 dm ^{3} |

9 | What volume of ethyne is evolved by reacting with 2.5 moles of calcium carbide with water CaC_{2} + 2H_{2}O ==Ca(OH)_{2}+ C_{2}H_{2} | CaC _{2} + 2H_{2}O ==Ca(OH)_{2}+ C_{2}H_{2}1 mole of CaC=== 1 mole of _{2} C _{2}H_{2}2.5 moles of CaC=== x _{2}X = 0.95 x 1/4 = 0.2375 mole of O _{2} Then we convert the mole to mass Since I mole of O _{2}=== 22.4 dm 0.2375 mole of O _{2}=== x X = 0.2375 X 22.4 = 5.249 dm ^{3} |

MASS TO MASS STOICHIOMETRY | ||

10 | What mass of carbon monoxide is evolved by 20g of carbon burning in oxygen 2C + O_{2} ===2CO | 2C + O 2 moles of _{2} ===2COC === 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 N _{2} + 3H_{2} ==2NH_{3}_{ } | N _{2} + 3H_{2} ==2NH_{3}1 mole of N=== 2 moles of _{2} NH _{3} Converting to mass 28 grams of N _{2}=== 34 grams of NH_{3} 30 grams of N _{2} === 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? CaCO_{3} ==CaO + CO_{2} | CaCO _{3} ==CaO + CO_{2}1 mole of CaCO=== 1 moles of _{3} CO _{2} Converting to mass 100 grams of CaCO _{3}=== 44 grams of CO_{2} 58 grams of CaCO _{3}=== 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 2KHCO_{3} ===K_{2}CO_{3} + H_{2}O +O_{2} | 2KHCO 2 moles of _{3} ===K_{2}CO_{3} + H_{2}O +O_{2}KHCO=== 1 mole of O_{3} _{2} 2 X100 grams of KHCO_{3}=== 32 grams of O_{2} 200 grams of KHCO_{3}=== 32 grams of O_{2} 29 grams of KHCO_{3}=== x X = 29 x 32 / 200 X = 4.64 grams of Oxygen |

14 | 18.8 grams of Nitrogen reacted with Hydrogen N _{2} + 3H_{2} ===2NH_{3}Calculate the volume of ammonia evolved. | N _{2} + 3H_{2} ===2NH_{3} 1 mole of N=== 2 mole of NH_{2} _{3} 28 grams of N _{2}=== 44.8 dm^{3} of NH_{3} 18.8 grams of N _{2}=== x X = 30.08 dm ^{3} of Ammonia |

VOLUME TO VOLUME STOICHIOMETRY | ||

15 | What volume of sulphur vi oxide is evolved by reacting 12.6 dm ^{3} of sulphur iv oxide.2SO_{2} + O_{2} ==2SO_{3} | 2SO_{2} + O_{2} ==2SO_{3}2 moles of SO=== 2 moles of SO_{2}_{3} Converting moles to volume 44.8 dm ^{3} of SO_{2}=== 44.8 dm^{3} of SO_{3} 12.6 dm ^{3} of SO_{2}=== x X = 12.6 x 44.8 /44.8 X = 12.6 dm ^{3} |

16 | What volume of carbon iv oxide is evolved by burning 300cm ^{3}of methane?CH_{4} + 2O_{2}===CO_{2} +2H_{2}O | CH _{4} + 2O_{2}===CO_{2} +2H_{2}O1 moles of CH=== 1 mole of CO_{4}_{2} Converting moles to volume 22.4 dm ^{3} of CH_{4}=== 22.4 dm^{3} of CO_{2} 300 cm ^{3} means 3 dm^{3} of CH_{4}3 dm ^{3} of CH_{4} === x X = 3 x 22.4 /22.4 X = 3 dm ^{3} |

So, stoichiometry table worksheet above illustrates the different steps and practice questions necessary in solving stoichiometry problems.