Question

If i combined 15.0 grams of calcium hydroxide with 75.0 ml of 0.500 m hcl, how many grams of calcium chloride would be formed?

Answer 1

To find the grams of calcium chloride formed, we need to determine the limiting reactant and use stoichiometry to convert from moles to grams.

Step-by-step explanation:

To calculate the grams of calcium chloride formed when 15.0 grams of calcium hydroxide reacts with 75.0 mL of 0.500 M HCl, we need to determine the limiting reactant. First, we need to convert the volume of HCl to moles by multiplying 75.0 mL by the molarity (0.500 mol/L). This gives us 0.0375 moles of HCl. Next, we convert the grams of calcium hydroxide to moles by dividing 15.0 grams by the molar mass of calcium hydroxide (Ca(OH)2). The molar mass of Ca(OH)2 is 74.10 g/mol, so we have 0.202 moles of Ca(OH)2.

Based on the balanced chemical equation Ca(OH)2 + 2HCl → CaCl2 + 2H2O, we can see that one mole of calcium hydroxide reacts with two moles of HCl to form one mole of calcium chloride. Therefore, the moles of calcium chloride formed is equal to half the moles of HCl, which is 0.0188 moles.

Finally, we can convert the moles of calcium chloride to grams by multiplying by the molar mass of calcium chloride (110.98 g/mol). This gives us 2.09 grams of calcium chloride formed.

Answer 2

The equation is:
Ca(OH)₂(s) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂O(l)
n=mass in g/M.M15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂no. of mol of HCl:n=0.5 mol/L*0.075L=0.0375 molThis could react with 0.0375/2= 0.01875 mol of Ca(OH)₂ We have a lot more than that.Therefore, HCl is the limiting reagent and determines how much CaCl₂ forms.Based on the balanced reaction, 2 moles of HCl gives 1 mole of CaCl₂no. of mol of CaCl₂= 0.0375/2= 0.01875 molmass in g=n*MM= 0.01875*111= 2.08 g