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?

n=mass in g/M.M

15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂

no. of mol of HCl:

n=0.5 mol/L*0.075L=0.0375 mol

This 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 CaCl₂

no. of mol of CaCl₂= 0.0375/2= 0.01875 mol

mass in g=n*MM= 0.01875*111= 2.08 g

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

the answers would be

the first one and the second one