Stoichiometry And Stoichiometric Calculations

The reaction between

%5Ctext%7BCaCO%7D_3

and

%5Ctext%7BHCl%7D

is as follows:

%5Ctext%7BCaCO%7D_3%20%2B%202%20%5C%2C%20%5Ctext%7BHCl%7D%20%5Crightarrow%20%5Ctext%7BCaCl%7D_2%20%2B%20%5Ctext%7BCO%7D_2%20%2B%20%5Ctext%7BH%7D_2%5Ctext%7BO%7D

### Step 1: Find the Molar Mass of

%5Ctext%7BCaCO%7D_3

The molar mass of

%5Ctext%7BCaCO%7D_3

is:

%5Ctext%7BMolar%20mass%20of%20%7D%20%5Ctext%7BCaCO%7D_3%20%3D%2040%20%2B%2012%20%2B%20(3%20%5Ctimes%2016)%20%3D%20100%20%5C%2C%20%5Ctext%7Bg%2Fmol%7D

### Step 2: Find the Moles of

%5Ctext%7BCaCO%7D_3

Given that 10 g of

%5Ctext%7BCaCO%7D_3

is used and it is 90% pure, the effective mass of

%5Ctext%7BCaCO%7D_3

is:

%5Ctext%7BMass%20of%20pure%20%7D%20%5Ctext%7BCaCO%7D_3%20%3D%2010%20%5Ctimes%200.90%20%3D%209%20%5C%2C%20%5Ctext%7Bg%7D

Now, calculate the moles of

%5Ctext%7BCaCO%7D_3

%5Ctext%7BMoles%20of%20%7D%20%5Ctext%7BCaCO%7D_3%20%3D%20%5Cfrac%7B9%7D%7B100%7D%20%3D%200.09%20%5C%2C%20%5Ctext%7Bmol%7D

### Step 3: Moles of

%5Ctext%7BCO%7D_2

From the balanced equation, 1 mole of

%5Ctext%7BCaCO%7D_3

produces 1 mole of

%5Ctext%7BCO%7D_2

. Therefore, the moles of

%5Ctext%7BCO%7D_2

produced are also 0.09 mol. ### Step 4: Find the Mass of

%5Ctext%7BCO%7D_2

The molar mass of

%5Ctext%7BCO%7D_2

is:

%5Ctext%7BMolar%20mass%20of%20%7D%20%5Ctext%7BCO%7D_2%20%3D%2012%20%2B%20(2%20%5Ctimes%2016)%20%3D%2044%20%5C%2C%20%5Ctext%7Bg%2Fmol%7D

Thus, the mass of

%5Ctext%7BCO%7D_2

is:

%5Ctext%7BMass%20of%20%7D%20%5Ctext%7BCO%7D_2%20%3D%200.09%20%5Ctimes%2044%20%3D%203.96%20%5C%2C%20%5Ctext%7Bg%7D

So, the mass of

%5Ctext%7BCO%7D_2

produced is approximately

4.4%20%5C%2C%20%5Ctext%7Bg%7D

. Thus, the correct answer is:

%5Cboxed%7B(A)%20%5C%2C%204.4%20%5C%2C%20%5Ctext%7Bg%7D%7D

Chapter Questions
4 MCQs