ch 9 rev ans
bs lab
• Write the equation for the reaction in the baking soda lab. Include states.
NaHCO3(s) + HCl(aq) => CO2(g) + H2O(l) + NaCl(aq)
Data: E dish =41.5g , e dish and bs =48.5g, e dish and salt =45.5g.
Find the theoretical yield of salt and %yield. Show work.
7g bs/1 * 1mole bs/84gbs * 1molesalt/1molebs * 58gsalt/1mole salt = 4.83g (salt theoretical)
Actual yield (in dish) 45.5-41.5=4g
%yield = actual/theoretical * 100 = 4/4.83 = 83%
Give the common names for the substances.
baking soda plus hydrochloric acid = carbon dioxide, water and saltwater
What are the products? carbon dioxide, water and saltwater.
reactants?baking soda plus hydrochloric acid.
Limiting reactant? bs
Excess reactant? HCl
What was in the dish when the reaction was over, before putting it into the oven?
saltwater
Which substance appeared as bubbles?
carbon dioxide
Which substance was removed by the oven?
water
Which substance was the residue after heating?
salt
Was this reaction endo or exothermic? Why?
endothermic; it got colder
About what temperature is the room in celsius?
22-23
mgo lab
• Write the equation for the reaction in the burning magnesium lab. Include states.
2Mg(s) + O2(g) =>(heat above arrow) 2MgO(s)
data: Crucible =12.3g, crucible and Mg =13.4g crucible and MgO = 13.8g.
Find the theoretical yield of product and %yield. Show work.
1.1gMg/1 * 1mole Mg/24gMg * 2mole MgO/2mole Mg * 40gMgO/1moleMgO = 1.83g theoretical yield MgO
actual yield 13.8 - 12.3 = 1.5gMgO
% yield = actual/theoretical * 100 = 1.5/1.83 = 82%
What are the products?
MgO.
reactants?
Mg and Oxygen.
Limiting reactant?
Mg.
Excess reactant?
oxygen.
What was the white smoke?
MgO particles suspended in air
Which substance had a molar mass of 24?
Mg
In this reaction it appears matter is__
created; it gains mass
When wood burns, it appears matter is __
destroyed; ashes weigh less than wood
In chemical changes matter is always ____
conserved. matter can not be created or destroyed
Was this reaction endo or exothermic? Why?
exothermic; heat is given off
Classify this reaction as one of the 5 types.
synthesis or combustion
What is the molar mass of magnesium oxide?
40g
stoichiometry
• Balance and classify:
decomposition 2KClO3 = 2 KCl + 3O2
How many moles of KCl are produced when 15 moles of potassium chlorate are decomposed?
15
How many moles of oxygen are produced from the decomposition of 1 mole of potassium chlorate?
1.5
How many moles of potassium chlorate are needed to make 12 moles of oxygen?
8
How many moles of KCl are produced when 3 moles of oxygen are obtained?
2
How many Liters of oxygen gas would be produced from 1 mole of potassium chlorate? (at STP)
1moleKClO3 / 1* 3mole O2/ 2moleKClO3 * 22.4L O2/ 1mole O2 =33.6L oxygen
How many Liters of oxygen gas (O2) are produced from the decomposition of 25g of Potassium chlorate?
25gKClO3 / 1 * 1moleKClO3 / 122gKClO3 /1 * 3mole O2/ 2moleKClO3 * 22.4L O2/ 1mole O2 = 6.9 L O2
combustion & stochiometry
What is the mass of 1 mole of C2H6?
30g
What is the volume of 1 mole of C2H6 at STP?
22.4L
What do the letters STP mean?
standard temperature
(0C)and pressure(1atm);
Write a balanced chemical equation for the combustion of heptane - C7H16
C7H16(g) + 11O2 (g) => (heat above arrow) 7CO2(g) + 8H2O(g)
What mass of carbon dioxide would be theoretically produced from the combustion of 25g of heptane? If the actual yield is 71g, what is the % yield? Show work
25gC7H16 /1 * 1moleC7H16 / 100gC7H16 * 7moleCO2/ 1moleC7H16 * 44gCO2 / 1mole CO2 = 77g carbon dioxide
ac/th= 71/77 = 92%
What are the products?
carbon dioxide and water.
reactants? heptane and oxygen.
Which reactant is always present in combustion?
oxygen.
Which is the fuel?
heptane
What is always produced when hydrocarbons burn?
carbon dioxide gas and water vapor
What is the match or striker needed to start the reaction called?
activation energy
What is the mass of ½ mole of water?
9g
What is the mass of 2 moles of carbon dioxide?
88g
Only HChem below this line ____________________________________________________
balance the equation, circle the limiting reactant, fill in the table
1mol C2H6/1 * 7mole O2/2mole C2H6 = 3.5 mole O2
Write a balanced chemical equation for the combustion of C6H14.
2C6H14 (g) + 19O2 (g) => (heat above arrow) 12CO2(g) + 14H2O(g)
40g of oxygen gas (1.25moles) and 20 g (.23 mole) of the C6H14 are mixed and ignited. What is the limiting reactant? How many grams of water are produced?
20g C6H14 * 1mole of C6H14/ 86g * 14mole water/2mole C6H14. * 18g/1mole water = 29.3 gwater
40gox * 1mole ox/ 32g * 14mole water/19mole ox * 18g/1mole water = 16.6gwater
therefore, ox is limiting reactant and 16.6g of water are formed. (29.3g can not be formed because reaction stopped when ox ran out)
How many grams of excess reactant remain?
40gox * 1mole ox/ 32g * 2mole c6/19mole ox *86g/1mol c6 = 11.3g are needed, therefore 20-11.3 = 8.7g of excess reactant (C2H6)remains.
#26
3H2SO4 + 2Al(OH)3 => 6H2O + Al2(SO4)3
to find lr:
30g H2SO4 /1 * 1mol H2SO4 / 98g H2SO4 * 6mol water/ 3molH2SO4 * 18g water/1mole = 11g water
25g aluminum hydroxide/1 * 1mol / 99g * 6mol water/2mol aluminum hydroxide * 18g/1mole= 13.6g water
therefor, lr is sulfuric acid and 11g water are formed
other product = 34.9 g aluminum sulfate
9.05g of excess reactant remains.