Roles Player 1 set up problem and work through first conversion factor Player 2 work through second conversion factor if there is one Player 3 work through third conversion factor if there is one ID: 816606
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Slide1
Stoichiometry Board Relay
Slide2Roles:
Player 1
:
set up problem and work through first conversion factor
Player 2
:
work through second conversion factor (if there is one)
Player 3
:
work through third conversion factor (if there is one)
Player 4
:
check work and
solve If there is no 4
th
person in your group, All students will check work and solve.
Slide3Write and balance the following reaction:
Solutions of sodium carbonate and barium
nitrate react. Predict the products.
Slide41
Na
2
CO
3
+ 1 Ba(NO3)2 1 BaCO3 + 2 NaNO3
Solutions of sodium carbonate and barium nitrate react. Predict the products.
Slide5Given 0.50 moles of barium nitrate, how many moles of sodium nitrate are formed?
Slide6Given 0.50 moles of barium nitrate, how many moles of sodium nitrate are formed?
1
Na
2
CO
3 + 1 Ba(NO3)2 1 BaCO3 + 2 NaNO3
0.50
mol
Ba(NO3)2 2 mol NaNO3 = 1 mole NaNO3 1 mol Ba(NO3)2
Slide7What mass of barium carbonate will form if 25.0g of sodium carbonate reacts with excess barium nitrate?
Slide8What mass of barium carbonate will form if 25.0g of sodium carbonate reacts with excess barium nitrate
?
1
Na
2
CO3 + 1 Ba(NO3)2 1 BaCO3 + 2 NaNO325.0g Na2CO3 1 mole Na2CO
3
1 mol BaCO3 197.3g BaCO3 106g Na2CO3 1 mol Na2CO3
1 mol
BaCO3
= 46.5g
BaCO3
Slide9What mass of
sodium nitrate will
form if
3.2 x 10
24
formula units barium nitrate reacts with excess sodium carbonate?
Slide10What mass of sodium nitrate will form if 3.2 x 10
24
formula units barium nitrate reacts with excess sodium carbonate?
1
Na2CO3 + 1 Ba(NO3)2 1 BaCO3 + 2 NaNO33.2 x 1024 f.u
.
Ba(NO
3)2 1 mole Ba(NO3)2 2 mol NaNO3 85g 6.02 x 1023 f.u. Ba(NO3)2 1
mol Ba(NO3)2
1
mol
= 903.7g NaNO
3
Slide11Write and balance the following reaction:
Lithium and nitrogen gas react. Predict the products.
Slide126
Li +
1
N
2 2 Li3NLithium and nitrogen gas react. Predict the products.
Slide13What mass of lithium is needed to react with 15.0g of nitrogen?
Slide14What mass of lithium is needed to react with 15.0g of nitrogen?
6
Li +
1
N
2 2 Li3N15.0g N2 1 mol N2 6 mol Li 6.9g Li 28.0g N2
1
mol N2 1 mol Li = 22.2g Li
Slide15What mass of lithium is needed to form 12.2g of lithium nitride?
Slide16What mass of lithium is needed to form 12.2g of lithium nitride?
6
Li +
1
N
2 2 Li3N12.2g Li3N 1 mol Li3N 6 mol Li 6.9g Li 34.7g Li3N 2
mol
Li3N 1 mol Li = 7.3g Li
Slide17How many atoms of Lithium react with 2.8 x 10
22
molecules of nitrogen?
Slide18How many atoms of Lithium react with 2.8 x 10
22
molecules of nitrogen?
6
Li +
1 N2 2 Li3N2.8 x 1022 m/c N2 1 mol N2 6 mol
Li 6.02
x 10
23 at Li 6.02 x 1023 m/c N2 1 mol N2 1 mol Li = 4.7 x 1021 atoms Li
Slide19Write and balance the following reaction:
Solutions of magnesium sulfate and aluminum nitrate are
mixed. Predict the products.
Slide203
MgSO
4
+
2
Al(NO3)3 3 Mg(NO3)2 + 1 Al2
(SO
4
)3Solutions of magnesium sulfate and aluminum nitrate are mixed. Predict the products.
Slide21How many grams of magnesium sulfate are needed to react with 0.75 moles of aluminum nitrate?
Slide22How many grams of magnesium sulfate are needed to react with 0.75 moles of aluminum nitrate
?
3
MgSO
4
+ 2 Al(NO3)3 3 Mg(NO3)2 + 1 Al2(SO4)
3
0.75
mol Al(NO3)3 3 moles MgSO4 120.4g MgSO4 2 moles Al(NO3)3 1 mol MgSO4 = 135.5g MgSO4
Slide23What mass of aluminum sulfate will form if 12.0g of aluminum nitrate react with 20.5g of magnesium sulfate
? Identify the limiting reactant and excess reactant.
Slide24What mass of aluminum sulfate will form if 12.0g of aluminum nitrate react with 20.5g of magnesium sulfate? Identify the limiting reactant and excess reactant
.
3
MgSO
4
+ 2 Al(NO3)3 3 Mg(NO3)2 + 1 Al2(SO4)3
12.0g Al(NO
3
)3 1 mol Al(NO3)3 1 mol Al2(SO4)3 342.3g Al2
(SO4)3
213.0g Al(NO3)3
2
mol Al(NO3)3 1
mol Al2(SO
4)3
= 9.6g Al2(SO
4)3
20.5g
MgSO
4
1
mol
MgSO
4
1
mol
Al
2
(SO
4
)
3
342.3
g
Al
2
(SO
4
)
3
120.4g
MgSO
4
3
mol
MgSO
4
1
mol
Al
2
(SO4)3 = 19.4g Al2(SO4)3L.R. = Al(NO3)3 E.R. = MgSO4
Slide25Write and balance the following reaction:
Write and balance the equation for the combustion of
dicarbon
dihydride
.
Slide262
C
2
H
2
+ 5 O2 4 CO2 + 2 H2O
Write and balance the equation for the combustion of
dicarbon
dihydride.
Slide27How many moles of water will be produced if 3.8 moles of oxygen react with excess
dicarbon
dihydride
?
Slide28How many moles of water will be produced if 3.8 moles of oxygen react with excess
dicarbon
dihydride
?
2 C2H2 + 5 O2 4 CO2 + 2 H2O
3.8
mol
O2 2 moles H2O = 1.5 moles H2O 5 moles O2
Slide29How many molecules of carbon dioxide will be produced if 0.85 moles of water are produced?
Slide30How many molecules of carbon dioxide will be produced if 0.85 moles of water are produced?
2
C
2
H
2 + 5 O2 4 CO2 + 2 H2O 0.85
mol
H2O 4 moles CO2 6.02 x 1023 m/c CO2 2 moles O2 1 mol CO2 = 1.0
x 1024
m/c CO2
Slide31If 5.8 grams of
dicarbon
dihydride
react with 4.2 grams of oxygen gas, how many grams of water will be produced? Identify the limiting reactant and what is the excess reactant.
Slide32If 5.8 grams of
dicarbon
dihydride
react with 4.2 grams of oxygen gas, how many grams of water will be produced? Identify the limiting reactant and what is the excess reactant.
2 C2H2 + 5 O2 4 CO2 + 2 H2O
5.8g
C
2H2 1 mole C2H2 2 moles H2O 18.0 g H2O 26.0g C2H2 2 moles C2H
2 1 mole H2O
= 4.0g H2O
4.2g
O
2 1 mole
O2 2
moles H2O 18.0
g H2O
32.0g O2 5 moles O
2
1
mole
H
2
O
= 0.9 g
H
2
O
L.R. = O
2
E.R. = C
2
H
2