A27.86 ml sample of 0.1744 m hno3 is tirades with 29.4ml of a job solution. what is the molarity of the koh

0.165 mol·L⁻¹

Explanation:

1. Write the chemical equation for the reaction.

HNO₃ + KOH ⟶ KNO₃ + H₂O

2. Calculate the moles of HNO₃

c = n/V                                               Multiply each side by V and transpose

n = Vc

V = 0.027 86 L

c = 0.1744 mol·L⁻¹                             Calculate the moles of HNO₃

Moles of HNO₃ = 0.027 86 × 0.1744

Moles of HNO₃ = 4.859 × 10⁻³ mol HNO₃

3. Calculate the moles of KOH

1 mol KOH ≡ 1 mol HNO₃                 Calculate the moles of KOH

Moles of KOH = 4.859 × 10⁻³× 1/1

Moles of KOH = 4.859 × 10⁻³ mol KOH

4. Calculate the molar concentration of the KOH

V = 29.4 mL = 0.0294 L                   Calculate the concentration

c = 4.859 × 10⁻³/0.0294

c = 0.165 mol·L⁻¹

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"