For each of the reactions, calculate the mass (in grams) of the product formed when 3.68 g of the underlined reactant completely reacts. assume that there is more than enough of the other reactant.
part a: ba(s)+cl₂(g) bacl₂(s)
part b: cao(s)+co₂(g) caco₃(s)
part c: 2mg(s)+o₂(g) 2mgo(s)
part d: 4al(s)+3o₂(g) 2al₂o₃(s)

As it is not clear which element is underlined there are to options per each part.

Part A

Option 1:  if we consider underlined the Ba(s),  5.62 g BaCl2(s) are formed.  

Option 2: if we consider underlined the Cl2(g) , 10.83 g BaCl2(s) are formed.  

Part B

Option 1:  if we consider underlined the CaO (s), 6.61 g CaCO3(s) are formed.  

Option 2: if we consider underlined the CO2 (g) , 8.41 g CaCO3(s) are formed.        

Part C

Option 1:  if we consider underlined the Mg (s),  6.08 g MgO(s) are formed.  

Option 2: if we consider underlined the O2 (g) , 9.26  g MgO(s) are formed.   .  

Part D

Option 1:  if we consider underlined the CaO (s), 3.46 g Al2O3 (s) are formed.  

Option 2: if we consider underlined the CO2 (g) ,  3.91 g Al2O3 (s) are formed.  .  

Explanation:

Part A

Ba(s) +Cl2(g) ⇒ BaCl2(s)

Step 1: We check if the reaction is well balanced, we count the atoms of the rectives (left) and we count the atoms of the producs (right)  

Reactives Products  

1 Ba 1

2 Cl 2  

Now that we have proved the reaction is well balanced we proceed to calculate.  

Option 1 , if we consider underlined the Ba  

3.68g Ba (1mol Ba/137.33 g Ba) = 0.027 mol Ba  

Because the relation in the balanced ecuation is 1, we can say that per 1 mol of Ba we obtain 1 mol of BaCl2, here we have 0.027 mol of Ba

0.027 mol Ba (1 mol BaCl2/ 1mol Ba) (208.23g BaCl2/ 1 mol BaCl2) = 5.62 g BaCl2 are formed.  

Option 2, if we consider underlined the Cl2  

3.68g Cl2 (1mol Cl2/70.9 g Cl2) = 0.052 mol Ba  

Because the relation in the balanced ecuation is 1, we can say that per 1 mol of Cl2 we obtain 1 mol of BaCl2, here we have 0.052 mol of Cl2

0.052 mol Cl2 (1 mol BaCl2/ 1mol Cl2) (208.23g BaCl2/ 1 mol BaCl2) = 10.83 g BaCl2 are formed.  

Part B

CaO(s) +CO2(g) ⇒CaCO3(s)

Step 1: We check if the reaction is well balanced, we count the atoms of the rectives (left) and we count the atoms of the producs (right)  

Reactives Products  

1 Ca 1

3 O 3  

1 C 1

Now that we have proved the reaction is well balanced we proceed to calculate.  

3.68g CaO (1mol CaO/56.077 g CaO) = 0.066 mol CaO  

Because the relation in the balanced ecuation is 1, we can say that per 1 mol of CaO we obtain 1 mol of CaCO3, here we have 0.066 mol of Ba

0.066 mol CaO (1 mol CaCO3/ 1mol CaO) (100.085g CaCO3/ 1 mol CaCO3) = 6.61 g CaCO3 are formed.  

3.68g CO2 (1mol CO2/44 g CO2) = 0.084 mol Ba  

Because the relation in the balanced ecuation is 1, we can say that per 1 mol of CO2 we obtain 1 mol of CaCO3, here we have 0.084 mol of CO2

0.084 mol CO2 (1 mol CaCO3/ 1mol CO2) (100.085 g CaCO3/ 1 mol CaCO3) = 8.41 g CaCO3 are formed.  

Part C

2Mg (s) +O2(g) ⇒ 2MgO(s)

Step 1: We check if the reaction is well balanced, we count the atoms of the rectives (left) and we count the atoms of the producs (right)  

Reactives Products  

2 Mg 2

2 O 2  

Now that we have proved the reaction is well balanced we proceed to calculate.  

3.68g Mg (2mol Mg /48.61 g Mg) = 0.151 mol Mg  

Because the relation in the balanced ecuation is 2 for Mg, we can say that per 2 mol of Mg we obtain 2 mol of MgO, here we have 0.151 mol of Mg

0.151 mol Mg (2 mol MgO/ 2 mol Mg) (80.60 g MgO/ 2 mol MgO) = 6.08 g MgO are formed.  

3.68g O2 (1mol O2 /32 g O2) = 0.115 mol O2  

Because the relation in the balanced ecuation is 1 for O2, we can say that per 1 mol of O2 we obtain 2 mol of MgO, here we have 0.115 mol of O2

0.115 mol O2 (2 mol MgO/ 1 mol O2) (80.60 g MgO/ 2 mol MgO) = 9.26  g MgO are formed.  

Part D

4 Al (s) +3O2(g) ⇒ 2Al2O3(s)

Step 1: We check if the reaction is well balanced, we count the atoms of the rectives (left) and we count the atoms of the producs (right)  

Reactives Products  

4 Al 4

6 O 6  

Now that we have proved the reaction is well balanced we proceed to calculate.  

3.68g Al (1 mol Al / 27.98 g Al) = 0.136 mol Al  

Because the relation in the balanced ecuation is 4 for Al, we can say that per each mol of Al we obtain 2 mol of Al2O3, here we have 0.136 mol of A

0.136 mol Al (2 mol Al2O3/ 4 mol Al) (102 g Al2O3/ 1 mol Al2O3) = 6.94 g Al2O3 are formed.  

3.68g O2 (3mol O2 /96 g O2) = 0.115 mol O2  

Because the relation in the balanced ecuation is 3 for O2, we can say that per 3 mol of O2 we obtain 2 mol of Al2O3, here we have 0.115 mol of O2

0.115 mol O2 (2 mol Al2O3/ 3 mol O2) (102 g Al2O3/ 2 mol Al2O3) = 3.91 g Al2O3 are formed.  

10 years is the answer