When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
CaCO3 (s) + 2HCL (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)

How many grams of calcium chloride will be produced when 25.0 g of calcium carbonate is combined with 10.0 g of hydrochloric acid?

Which reactant is in excess? HCl or CaCO3?

How many grams of the excess reactant will remain after the reaction is complete?

Mass of CaCl₂ produced = 15 g

Excess reactant = CaCO₃

Mass of CaCO₃ left = 11.5 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 10.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  → CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:  

Number of moles of HCl = Mass /molar mass

Number of moles of HCl = 10.0 g / 36.5 g/mol

Number of moles of HCl = 0.27 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃      :               CaCl₂

                   1               :               1

                 0.25           :            0.25

                HCl              :                CaCl₂

                 2                :                 1

                 0.27            :               1/2 × 0.27 = 0.135 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.135 mol × 110.98 g/mol

Mass of CaCl₂ =  15 g

The calcium carbonate is present in excess.

               HCl              :                CaCO₃

                 2                :                 1

                 0.27            :               1/2 × 0.27 = 0.135 mol

So, 0.135 moles react with 0.27 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 - 0.135

The moles of CaCO₃ remain unreacted = 0.115 mol

Mass remain unreacted:

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.115 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted = 11.5 g

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )