Question

Not a timed or graded assignment. Please show all calculation work. Question prompt “what mass of the excess reactant HCI will be used”. Quick answer showing work = amazing review :)

Answer 1

1) Balance the chemical equation.

`CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2`

2) Convert grams to moles

The molar mass of CaCO3 is 100.086 g/mol

`molCaCO_3=10.0gCaCO_3*(1molCaaCO_3)/(100.086gCaCO_3)=0.0999molCaCO_3`

We have 0.0999 mol CaCO3.

The molar mass of HCl is 36.46 g/mol

`molHCl=15.0gHCl*(1molHCl)/(36.46gHCl)=0.411molHCl`

We have 0.411 mol HCl.

3) Which is the limiting reactant?

How many moles of CaCO3 do we need to use all of the HCl?

The molar ratio is 1 mol CaCO3: 2 mol HCl.

`molCaCO_3=0.411molHCl*(1mnolCaCO_3)/(2molHCl)=0.2055molCaCO_3`

We have 0.0999 mol CaCO3 and we need 0.2055 mol CaCO3. We do not have enough CaCO3. This is the limiting reactant.

How many moles of HCl do we need to use all of the CaCO3?

The molar ratio is 1 mol CaCO3: 2 mol HCl.

`molHCl=0.0999molCaCO_3*(2molHCl)/(1molCaCO_3)=0.1998molHCl`

We need 0.1998 mol HCl and we have 0.411 mol HCl. We have enough HCl. This is the excess reactant.

4) What mass of the excess reactant will be used?

The excess reactant is HCl because we have 0.411 mol and we just need 0.1998 mol. In the reaction, we will use 0.1998 mol HCl.

5) Convert grams to moles

The molar mass of HCl is 36.46 g/mol

`gHCl=0.1998molHCl*(36.46gHCl)/(1molHCl)=7.28gHCl`

7.28 g HCl will be used in the reaction.