Information given by chemical equations
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Information given by chemical equations. 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g). In this equation there are 2 molecules of benzene reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water .

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Information given by chemical equations

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Information given by chemical equations 1341762

Information given by chemical equations

2 C6H6(l) +15O2(g)12CO2(g) +6H2O (g)

  • In this equation there are 2 molecules of benzene reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water.

  • This equation could also be read as 2 moles of benzene reacts with 15 moles of oxygen to produce 12 moles of carbon dioxide and 6 moles of water.

Since the relationship between the actual number of molecules and the number of moles present is 6.02 x 1023, a common factor between all species involved in the equation, a MOLE RATIO relationship can be discussed.


Information given by chemical equations 1341762

Information given by chemical equations

2 C6H6(l) +15O2(g)12CO2(g) +6H2O (g)

The MOLE RATIO for benzene and oxygen is 2 : 15. It can be written as:

2 moles C6H6or as 15 moles of O2

15 moles O2 2 moles of C6H6

The MOLE RATIO for oxygen and carbon dioxide is 15 : 12. It can be written as:

12 moles CO2or as 15 moles of O2

15 moles O2 12 moles of CO2

  • NOTE:The MOLERATIO is used for converting moles of one substance into moles of another substance. Without the balanced equation there is no other relationship between two different compounds.


Information given by chemical equations 1341762

Using the mole ratio to relate the moles of one compound to the moles of another compound is the part of chemistry called

STOICHIOMETRY !!!!!

How many moles of oxygen are required to burn 2.40 moles of ethane, C2H6(g)

2 C2H6(g) + 7 O2(g)→ 4CO2(g) + 6 H2O (l)

2.40 moles C2H6(g) 7 moles O2 = 8.40 mole O2 (g)

2 mole C2H6(g)


Information given by chemical equations 1341762

STOICHIOMETRY !!!!!

2 H2(g)+ O2(g)2 H2O (g)

Q. How many mole of hydrogen are necessary to react with 2 moles of oxygen in order to produce exactly 4 moles of water?

A. 2 mol O2 (2 moles H2 / 1 mole O2) = 4 mole H2


Information given by chemical equations 1341762

STOICHIOMETRY

The Stoichiometry Flow Chart

Use mole ratio from equation

Use Molar mass (A)

Use Molar mass (B)


Information given by chemical equations 1341762

STOICHIOMETRY

2 H2(g) + O2(g) 2 H2O (g)

Q1. How many moles of hydrogen are necessary to react with 15.0 g of oxygen?

A. 15.0g O2(1 mole O2)(2 mole H2)= 0.938 moles H2

32.0 g 1 mole O2

Q2. How many grams of hydrogen are necessary to react with 15.0 g of oxygen?

A. 15.0g O2 (1 mole O2) ( 2 mole H2) ( 2.016 g H2) = 1.89 g H2

32.0 g 1 mole O21 mole H2


Information given by chemical equations 1341762

STOICHIOMETRY

2 H2(g) + O2(g) 2 H2O (g)

Q3. How many grams of water are produced from 15.0 g of oxygen?

A.15.0g O2 (1 mole O2) ( 2 mole H2O) ( 18.0 g H2O) =16.9 g H2O

32.0 g 1 mole O21 mole H2O

Q4. How much hydrogen and oxygen is needed to produce 25.0 grams of water?

A. 25.0g H2O (1 mole H2O)(2 mole H2)(2.016 g H2)= 2.80 g H2

18.0 g 2 mole H2O1 mole H2

A. 25.0g H2O (1 mole H2O)(1 mole O2)(32.0 g O2)= 22.2 g O2

18.0 g 2 mole H2O1 mole O2

Notice that the Law of Conservation of Mass still applies.


Information given by chemical equations 1341762

STOICHIOMETRY

Sn (s)+ 2Cl2 (g)SnCl4(s)

Q1. How many moles of chlorine gas are required to react with of 1.25 g of metallic tin to produce tin (IV) chloride?

A.1.25 g Sn(1 mole Sn)(2 mole Cl2)=0.0210 moles Cl2

119 g Sn 1 mole Sn

Q2. How many grams of tin (IV) chloride are formed from the reaction of 1.25 g of metallic tin and yellow chlorine gas?

A. 1.25 g Sn (1 mole Sn) ( 1 mole SnCl4) ( 261 g SnCl4) = 2.74 g SnCl4

119 g Sn 1 mole Sn1 mole SnCl4


Information given by chemical equations 1341762

How many grams of solid are formed when 10.0 g of lead reacts with excess phosphoric acid?

  • Write the chemical equation: Pb + H3PO4 ?

You recognize that this is a single displacement (replacement) reaction. So Pb (a metal) will displace (replace) H (the cation).

Pb + H3PO4 Pb3(PO4)2 + H2

2. Balance the equation: 3Pb+2 H3PO4 Pb3(PO4)2 + 3 H2

3. Make a list under the appropriate substance

3Pb+2 H3PO4 Pb3(PO4)2 (s) + 3 H2 (g)

10.0gm=?

Start with what is given:

10.0gPb (1 mole Pb)(1 mole Pb3(PO4)2)(811 g Pb3(PO4)2) = 13.1 g Pb3(PO4)2

207 g Pb 3 mole Pb1 mole Pb3(PO4)2


Practice problem 18

PRACTICE PROBLEM # 18

14.20g

1. How many grams of gas can be produced from 0.8876 moles of HgO?

2 HgO  2 Hg + O2

2. How many moles of fluorine are required to produce 12.0 grams of KrF6? Given the equation: Kr + 3 F2 KrF6

3. How many grams of Na2CO3 will be produced from the thermal decomposition of 250.0 g of NaHCO3?

4. How many grams of CO2 can be produced by the reaction of 75.0 grams of C2H2 with excess oxygen?

5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag.

How many grams of silver is produced when 125.0 g of copper is reacted with excess silver nitrate solution?

0.182 mol

157.7 g

254 g

424.9 g


Group study problem 18

GROUP STUDY PROBLEM # 18

______1. How many grams of liquid product can be produced from 3.55 moles of HgO?

2 HgO 2 Hg + O2

______2. How many moles of fluorine are required to produce 3.0 grams of KrF6? Given the equation: Kr + 3 F2 KrF6

______3. How many grams of Na2CO3 will be produced from the decomposition of 20.0g of NaHCO3?

______4. How many grams of O2 are needed to combust 55.0 grams of C2H4?

______5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag.

How many grams of silver is produced when 50.0 g of copper is reacted with excess silver nitrate solution?


Information given by chemical equations 1341762

STOICHIOMETRY

Molar Volume is the volume occupied by one mole of gas.

  • Molar volume is different than the molar mass because:

  • The molar mass is constant and independent of temperature and pressure. The molar volume of a gas is variable and is dependent on the temperature and pressure.

  • Each substance has its own unique molar mass. At a given temperature and pressure, all gases have the same molar volume

  • At standard temperature and pressure (STP), the molar volume of a gas is 22.4 L/mol.

  • One mole of any gas at STP occupies 22.4 liters.


Information given by chemical equations 1341762

Molar Volume Stoichiometry Problems

1. What volume is occupied by 4.21 moles of ammonia gas, NH3 at STP?

4.21 mol 22.4 L = 94.3 L

1 mole

2.What volume of hydrogen, measured at STP, can be released by 42.7 g of zinc as it reacts with hydorchloric acid?

Zn (s) + 2 HCl (aq)→ H2(g) + ZnCl2(aq)

42.7 g Zn 1 mol Zn1 mol H222.4 L H2 = 14.6 L H2

65.4 g Zn 1 mol Zn 1 mol H2


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