Stoichiometry Board Relay

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) Player 4: check work and solve If there is no 4th person in your group, All students will check work and solve.

Write and balance the following reaction: Solutions of sodium carbonate and barium nitrate react. Predict the products.

1 Na2CO3 + 1 Ba(NO3)2  1 BaCO3 + 2 NaNO3 Solutions of sodium carbonate and barium nitrate react. Predict the products. 1 Na2CO3 + 1 Ba(NO3)2  1 BaCO3 + 2 NaNO3

Given 0.50 moles of barium nitrate, how many moles of sodium nitrate are formed?

Given 0.50 moles of barium nitrate, how many moles of sodium nitrate are formed? 1 Na2CO3 + 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

What mass of barium carbonate will form if 25 What mass of barium carbonate will form if 25.0g of sodium carbonate reacts with excess barium nitrate?

What mass of barium carbonate will form if 25 What mass of barium carbonate will form if 25.0g of sodium carbonate reacts with excess barium nitrate? 1 Na2CO3 + 1 Ba(NO3)2  1 BaCO3 + 2 NaNO3 25.0g Na2CO3 1 mole Na2CO3 1 mol BaCO3 197.3g BaCO3 106g Na2CO3 1 mol Na2CO3 1 mol BaCO3 = 46.5g BaCO3

What mass of sodium nitrate will form if 3 What mass of sodium nitrate will form if 3.2 x 1024 formula units barium nitrate reacts with excess sodium carbonate?

1 Na2CO3 + 1 Ba(NO3)2  1 BaCO3 + 2 NaNO3 What mass of sodium nitrate will form if 3.2 x 1024 formula units barium nitrate reacts with excess sodium carbonate? 1 Na2CO3 + 1 Ba(NO3)2  1 BaCO3 + 2 NaNO3 3.2 x 1024 f.u. Ba(NO3)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 NaNO3

Write and balance the following reaction: Lithium and nitrogen gas react. Predict the products.

Lithium and nitrogen gas react. Predict the products. 6 Li + 1 N2  2 Li3N

What mass of lithium is needed to react with 15.0g of nitrogen?

What mass of lithium is needed to react with 15. 0g of nitrogen What mass of lithium is needed to react with 15.0g of nitrogen? 6 Li + 1 N2  2 Li3N 15.0g N2 1 mol N2 6 mol Li 6.9g Li 28.0g N2 1 mol N2 1 mol Li = 22.2g Li

What mass of lithium is needed to form 12.2g of lithium nitride?

What mass of lithium is needed to form 12. 2g of lithium nitride What mass of lithium is needed to form 12.2g of lithium nitride? 6 Li + 1 N2  2 Li3N 12.2g Li3N 1 mol Li3N 6 mol Li 6.9g Li 34.7g Li3N 2 mol Li3N 1 mol Li = 7.3g Li

How many atoms of Lithium react with 2.8 x 1022 molecules of nitrogen?

How many atoms of Lithium react with 2.8 x 1022 molecules of nitrogen? 6 Li + 1 N2  2 Li3N 2.8 x 1022 m/c N2 1 mol N2 6 mol Li 6.02 x 1023 at Li 6.02 x 1023 m/c N2 1 mol N2 1 mol Li = 4.7 x 1021 atoms Li

Write and balance the following reaction: Solutions of magnesium sulfate and aluminum nitrate are mixed. Predict the products.

3 MgSO4+ 2 Al(NO3)3  3 Mg(NO3)2 + 1 Al2(SO4)3 Solutions of magnesium sulfate and aluminum nitrate are mixed. Predict the products. 3 MgSO4+ 2 Al(NO3)3  3 Mg(NO3)2 + 1 Al2(SO4)3

How many grams of magnesium sulfate are needed to react with 0 How many grams of magnesium sulfate are needed to react with 0.75 moles of aluminum nitrate?

How many grams of magnesium sulfate are needed to react with 0 How many grams of magnesium sulfate are needed to react with 0.75 moles of aluminum nitrate? 3 MgSO4+ 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

What mass of aluminum sulfate will form if 12 What 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 MgSO4+ 2 Al(NO3)3  3 Mg(NO3)2 + 1 Al2(SO4)3 What 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 MgSO4+ 2 Al(NO3)3  3 Mg(NO3)2 + 1 Al2(SO4)3 12.0g Al(NO3)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(SO4)3 = 9.6g Al2(SO4)3 20.5g MgSO4 1 mol MgSO4 1 mol Al2(SO4)3 342.3g Al2(SO4)3 120.4g MgSO4 3 mol MgSO4 1 mol Al2(SO4)3 = 19.4g Al2(SO4)3 L.R. = Al(NO3)3 E.R. = MgSO4

Write and balance the following reaction: Write and balance the equation for the combustion of dicarbon dihydride.

Write and balance the equation for the combustion of dicarbon dihydride. 2 C2H2 + 5 O2  4 CO2 + 2 H2O

How many moles of water will be produced if 3 How many moles of water will be produced if 3.8 moles of oxygen react with excess dicarbon dihydride?

How many moles of water will be produced if 3 How 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

How many molecules of carbon dioxide will be produced if 0 How many molecules of carbon dioxide will be produced if 0.85 moles of water are produced?

How many molecules of carbon dioxide will be produced if 0 How many molecules of carbon dioxide will be produced if 0.85 moles of water are produced? 2 C2H2 + 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

If 5. 8 grams of dicarbon dihydride react with 4 If 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.

If 5. 8 grams of dicarbon dihydride react with 4 If 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 C2H2 1 mole C2H2 2 moles H2O 18.0 g H2O 26.0g C2H2 2 moles C2H2 1 mole H2O = 4.0g H2O 4.2g O2 1 mole O2 2 moles H2O 18.0 g H2O 32.0g O2 5 moles O2 1 mole H2O = 0.9 g H2O L.R. = O2 E.R. = C2H2