If 27.0 ml of ca(oh)2 with an unknown concentration is neutralized by 32.40 ml of 0.185 m hcl, what is the concentration of the ca(oh)2 solution? show all of the work needed to solve this problem. (2 points) ca(oh)2 + 2hcl yields 2h2 o + cacl2

Given information : Volume of HCl = 32.40 mL

Volume of HCl = 0.0324 L

Concentration of HCl = 0.185 M or 0.185 mol/L (M = mol/L)

Volume of Ca(OH)2 = 27.0 mL

Volume of Ca(OH)2 = 0.027 L

We need to find the concentration of Ca(OH)2.

To find the concentration of Ca(OH)2 we need moles and volume of Ca(OH)2.

Moles of Ca(OH)2 can be calculated using stoichiometry and volume of Ca(OH)2 is already given to us.

Step 1 : Find the moles of HCl using its given volume and concentration.

Moles of HCl = 0.005994 mol HCl

Step 2 : We need to find moles of Ca(OH)2 using mol of HCl with the help of mole ratio.

Mole ratio are the coefficient present in front of the compound in a balanced equation.

Mole ratio of Ca(OH)2 : HCl = 1:2 ( 1 coefficient of Ca(OH)2 and 2 coefficient of HCl)

Moles of Ca(OH)2 = 0.002997 mol Ca(OH)2

Step 3 : Find the concentration of Ca(OH)2 using its moles and volume.

Moles of Ca(OH)2 = 0.002997 mol and volume of Ca(OH)2 = 0.027 L

Concentration of Ca(OH)2 = 0.111 mol/L or 0.111 M

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.