CaCO3(s)+2HCl(aq)-->CaCl2(aq)+H2O(l)+CO2 (g)
How many grams of calcium chloride will be produced when 31g of calcium carbonate are combined with 11 g of HCl?
I have 0.3097 moles of CaCO3
and 0.30169 moles of HCl...I get stuck at that point
Look at the equation:
each mole of CaCO3 produces one mole of CaCl2.
So, you will wind up with .3097 moles of CaCl2.
Convert that to grams, and you're done.
CaCO3(s)+2HCl(aq)-->CaCl2(aq)+H2O(l)+CO2 (g)
0.3097 moles of CaCO3
0.30169 moles of HCl
0.3097 Moles CaCO3 X (1 Mole CaCl2 / 1 Mole CaCO3) = 0.3097 Mole CaCL2
0.30169 Moles HCL X ( 1 Mole CaCl2 / 2 Mole HCL) = 0.1508 Mole CaCl2
I think HCL is a limiting reactant: Therefore: 0.1508 Mole CaCl2 is the product amount.
To determine the grams of calcium chloride produced, we need to use the molar ratios from the balanced chemical equation.
1 mole of CaCO3 corresponds to 1 mole of CaCl2, so the moles of CaCl2 produced will be the same as the moles of CaCO3.
0.3097 moles of CaCO3 will produce 0.3097 moles of CaCl2.
Next, let's find the molar mass of CaCl2:
Ca: 1 atom x 40.08 g/mol = 40.08 g/mol
Cl: 2 atoms x 35.45 g/mol = 70.90 g/mol
Summing up the molar masses:
40.08 g/mol + 70.90 g/mol = 111.98 g/mol
Now, we can calculate the grams of CaCl2 produced:
0.3097 moles x 111.98 g/mol = 34.66 grams
Therefore, when 31 grams of calcium carbonate reacts with 11 grams of HCl, approximately 34.66 grams of calcium chloride will be produced.
To calculate the grams of calcium chloride produced in the reaction, we need to use the balanced equation and the molar masses of the compounds involved.
First, let's determine the molar mass of calcium chloride (CaCl2):
Ca: 1 atom x 40.08 g/mol = 40.08 g/mol
Cl: 2 atoms x 35.45 g/mol = 70.90 g/mol
Total molar mass: 40.08 + 70.90 = 110.98 g/mol
Next, we need to find out the limiting reactant. The limiting reactant is the one that limits the amount of product formed and is entirely consumed in the reaction. To determine the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
From your calculations, you have 0.3097 moles of CaCO3 and 0.30169 moles of HCl.
From the balanced equation, we can see that the stoichiometric ratio between CaCO3 and CaCl2 is 1:1. This means that for each mole of CaCO3, we produce 1 mole of CaCl2.
Since the moles of CaCO3 are greater than the moles of HCl, HCl is the limiting reactant in this case.
Now, let's calculate the moles of CaCl2 produced. According to the balanced equation, the stoichiometric ratio between HCl and CaCl2 is 2:1. This means that 2 moles of HCl react with 1 mole of CaCO3 to produce 1 mole of CaCl2.
Since the moles of HCl are 0.30169, this means that we have half that amount for the moles of CaCl2 produced: 0.30169 / 2 = 0.15085 moles.
Finally, we can calculate the grams of CaCl2 produced using the moles and molar mass:
Grams of CaCl2 = Moles of CaCl2 x Molar mass of CaCl2
Grams of CaCl2 = 0.15085 mol x 110.98 g/mol = 16.74 g
Therefore, when 31 g of calcium carbonate and 11 g of hydrochloric acid react, 16.74 grams of calcium chloride will be produced.