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.  

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).