CaCo3(s)+2HCl(aq)---->CaCl2(aq)+H2O(l)+CO2
How many grams of calcium chloride will be produced when 27.0g of calcium carbonate are combined with 14.0g of hydrochloric acid?
Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?
To find the number of grams of calcium chloride produced, we first need to determine the limiting reactant. A limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
Let's start by calculating the number of moles for both calcium carbonate (CaCO3) and hydrochloric acid (HCl) using their respective molar masses.
The molar mass of CaCO3 is:
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Total = 100.09 g/mol
Using the given mass of calcium carbonate, we can calculate the number of moles:
27.0 g CaCO3 × (1 mol CaCO3 / 100.09 g CaCO3) = 0.269 mol CaCO3
The molar mass of HCl is:
H = 1.01 g/mol
Cl = 35.45 g/mol
Total = 36.46 g/mol
Using the given mass of hydrochloric acid, we can calculate the number of moles:
14.0 g HCl × (1 mol HCl / 36.46 g HCl) = 0.384 mol HCl
Looking at the balanced equation, we can see that the stoichiometric ratio between CaCO3 and HCl is 1:2. This means that 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CaCl2.
However, we have 0.269 mol of CaCO3 and 0.384 mol of HCl. Since the ratio between CaCO3 and HCl is 1:2, we need twice as many moles of HCl as CaCO3 to react completely. So, we can conclude that CaCO3 is the limiting reactant and HCl is in excess.
To determine how much HCl remains after the reaction, we need to calculate the number of moles of HCl that reacted with CaCO3 and then subtract it from the initial moles of HCl.
Since 1 mole of CaCO3 reacts with 2 moles of HCl, and CaCO3 is limiting, the moles of HCl reacted can be calculated as:
0.269 mol CaCO3 × (2 mol HCl / 1 mol CaCO3) = 0.538 mol HCl (reacted)
Now, to find the remaining moles of HCl:
0.384 mol HCl (initial) - 0.538 mol HCl (reacted) = -0.154 mol HCl
Since the calculated value for remaining moles of HCl is negative, it means that none of the HCl will remain after the reaction is complete.
To calculate the mass of calcium chloride produced, we need to use the moles of limiting reactant (CaCO3) and its molar mass.
The molar mass of CaCl2 is:
Ca = 40.08 g/mol
Cl = 35.45 g/mol × 2 (two chloride ions present)
Total = 110.98 g/mol
Thus, the mass of calcium chloride produced is:
0.269 mol CaCO3 × (1 mol CaCl2 / 1 mol CaCO3) × 110.98 g/mol CaCl2 = 29.89 g CaCl2
Therefore, 29.89 grams of calcium chloride will be produced when 27.0 grams of calcium carbonate and 14.0 grams of hydrochloric acid react. HCl will be in excess, and no grams of HCl will remain after the reaction is complete.