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12 Weeks to TAKS Week 12. Objective 4: 8c: Law of Conservation of Mass The student is expected to investigate and identify the law of conservation of mass. Chemical Equations: An expression in which symbols and formulas are used to represent a chemical reaction.

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12 Weeks to TAKS Week 12


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    1. 12 Weeks to TAKSWeek 12

    2. Objective 4: 8c: Law of Conservation of Mass The student is expected to investigate and identify the law of conservation of mass.

    3. Chemical Equations:An expression in which symbols and formulas are used to represent a chemical reaction. sodium metal + chlorine gas  table salt (sodium chloride)

    4. The meaning of chemical equations A mathematical equation: x+2x=3x A chemical equation identifies the starting and finishing chemical as reactants and products: reactants products Example: Formation of water 2H2 + O2 2H20 A chemical equation is balanced when it reflects the conservation of mass and charge.

    5. Law of Conservation of Mass Mass is neither created nor destroyed during chemical or physical reactions. Total mass of reactants = Total mass of products Antoine Lavoisier

    6. The Law of conservation of massstates that matter cannot be created or destroyed in any chemical reaction The bonds between atoms in the reactants are rearranged to form new compounds, but none of the atoms disappear, and no new atoms are formed. So: Chemical equations must be balanced, meaning the numbers and kinds of atoms must be the same on both sides of the reaction arrow. The numbers placed in front of formulas to balance equations are called coefficients, and they multiply all the atoms in the chemical formula.

    7. Balancing Chemical Equations The following five steps can be used as a guide to balance chemical equations. Balance this chemical reaction. Sulfuric Acid reacts with sodium hydroxide to yield sodium sulfate and water Step 1: Write an unbalanced equation, using correct formulas for all reactants and products. H2SO4 + NaOH  Na2SO4 + H2O

    8. Step 2: Inventory all atoms found in the equation H2SO4 + NaOH  Na2SO4 + H2O 2 H + 1 H = 3H 2 Na 1 S 1 S 4 O + 1 O = 5 O 4 O + 1 O= 5 O 1 Na 2 H

    9. Equal Equal Step 3: Compare the number of each atom on each side of the equation. Add coefficients to balance the number of atoms. Remember that adding a coefficient affects all elements in the compound. H2SO4 + NaOH  Na2SO4 + H2O 2 H + 1 H = 3H 2 Na 1 S 1 S 4 O + 1 O = 5 O 4 O + 1 O= 5 O 1 Na 2 H 2 2 4H 6O 6 O 4H 2 Na

    10. Step 4: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are same. H2SO4 + 2 NaOH  Na2SO4 + 2 H2O 2 H + 2 H = 4H 2 Na 1 S 1 S 4 O + 2 O = 6 O 4 O + 2 O= 6 O 2 Na 4 H

    11. Step 5: Make sure the coefficients are reduced to their lowest whole-number value (ok here). H2SO4 + 2 NaOH Na2SO4 + 2 H2O 1 : 2: 1 : 2

    12. Your Turn! Balance the following equations. • KClO3 → KCl + O2 • P4 + O2 → P2O5 • Al2O3 → Al + O2 • Al2(SO4)3 + Ca(OH)2 → Al(OH)3 + CaSO4 • Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O

    13. Answers Balance the following equations. • 2 KClO3 → 2 KCl + 3O2 • P4 + 5O2 → 2 P2O5 • 2 Al2O3 →4 Al + 3O2 • Al2(SO4)3 +3 Ca(OH)2→ 2Al(OH)3 + 3CaSO4 • 3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O

    14. What do coefficients mean? • 1. They indicate the number of particles of atoms, molecules, and formula units found in the reaction • 2. They are used to determine the amount of reactants and products. 2 H2 + O2 2 H2O Indicates that 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of water.

    15. Before and After Reaction Particles always react in the same ratio and they are always conserved • 5 molecules + 15 molecules  10 molecules of N2of H2 of NH3 N2 + 3 H2 2 NH3

    16. Camels store the fat,tristearin (C57H110O6), in the hump. As well as being a source of energy, the fat is a source of water, because when it is used the reaction takes place. Given the following information, what mass of water can be made from 1000 g of fat? Water from a Camel fat 2 C57H110O6(s) + 163 O2(g)  114 CO2(g) + 110 H2O(l) 1000 g C57H110O6 + 2930 g O2 2817g CO2 + ? H2O

    17. Remember that mass is conserved. Therefore the mass of reactants = mass of products 1000 g C57H110O6 + 2930 g O2 2817g CO2 + ? H2O Mass of reactants = Mass of products = 3930 g – 2817g 1000 g + 2930 g = 1113 g H2O = 3930 g

    18. Your Turn! Water in Space In the space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. On average, each astronaut exhales about 880 g of CO2 daily. What mass of water will be produced when this amount of CO2 reacts with 956 g of LiOH according to the following equation? CO2(g) + 2 LiOH(s)  Li2CO3(aq) + H2O(l) 880 gCO2 + 956 g LiOH  1476 g Li2CO3 + ? H2O

    19. Remember that mass is conserved. Therefore the mass of reactants = mass of products 880 gCO2 + 956 g LiOH  1476 g Li2CO3 + ? H2O Mass of reactants = Mass of products = 1836 g - 1476 g 880 g + 956 g = 1836 g = 360 g H2O

    20. This demonstration is the combustion of diethyl ether in air. If 33.8 g of diethyl ether is added to the balloon, how many grams of carbon dioxide are given off? C2H5OC2H5 + 6 O2 4 CO2 + 5 H2O 33.8 g C2H5OC2H5 + 87.7 g O2 ? g CO2 + 41.2 g H2O = 80.3 g CO2

    21. If this experiment was done in your classroom, why would it be difficult to prove the law of conservation of mass? • CO2 and H2O are gases and they would move throughout the room. C2H5OC2H5 (l)+ 6 O2 (g)  4 CO2 (g) + 5 H2O (g)

    22. To produce 12 molecules of water, the flask must have how many molecules of ammonia (NH3)? 4 NH3 + 3 O22 N2+ 6 H2O 8 NH3 + 6 O2 4 N2 + 12 H2O (the ratio will still be 4:3:2:6) 8 molecules of NH3

    23. Methane gas is burned in excess oxygen to produce carbon dioxide & water. If 25.0 grams of methane is burned in 100. g of oxygen (O2) and 68.8 g CO2 are produced, how many grams of water is produced? CH4 (g) + 2 O2 (g)  2 H2O (l) + CO2 (g) 25.0 g + 100.0g = 68.8 g + ? 56.2 g H2O

    24. Phosphorus reacts with oxygen to produce diphosphorus pentoxide according to the equation: P4 (s) + 5 O2 (g)  2 P2O5 How many particles of phosphorus must be present to produce 30 molecules of P2O5? 15 P4 + 75 O2 30 P2O5 (ratio is still 1: 5: 2) 15 particles of P4