A mixture of sodium hydroxide and strontium hydroxide contains a total of 0.19 moles of the two compounds. It requires 100.0 mL of 2.969 M HCl to neutralize all of the base. How many moles of sodium hydroxide were in the original mixture

0.0831 moles of NaOH were in the original mixture

Explanation:

The strontium hydroxide, Sr(OH)2, requires 2 moles of HCl per mole of hydroxide to be neutralized.

The sodium hydroxide, NaOH, requires 1 mol of HCl per mol of hydroxide.

To solve this question we must write 2 equations:

(1) 0.19mol = X + Y

Where X = Moles NaOH; Y = Moles Sr(OH)2

The moles of HCl required are:

0.100L * (2.969mol / L) = 0.2969 moles HCl

0.2969 mol = X + 2Y(2)

Replacing (1) in (2):

0.19mol - Y = X

0.2969 mol = (0.19mol - Y) + 2Y

0.1069mol = Y

Moles X = Moles NaOH:

0.19mol = X + 0.1069mol

X = 0.0831 moles of NaOH were in the original mixture

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml