1. a) Calculate the moles of water produced when 0.1 mole of HCI reacts with 0.1

Question

1.

a) Calculate the moles of water produced when 0.1 mole of HCI reacts with 0.1 mole of NaOH. ( HCI + NaOH ---> H2O + NaCI )

b) If 0.5 moles of KOH are mixed with 1 mole of HNO3 how many moles of H2O are produced? ( KOH + HNO3 ---> KNO3 + H2O )

c) What would be the resulting temperature if 10 mL of water at 30 degrees Celsius is mixed with 100 mL of water at 90 degrees Celsius? Use the formula q=ms?t

d) One gallon of water at 0 degrees Celsius is mixed with 1 liter of water at 99.9 degrees Celsius. Using the formula q=ms?t calculate the resulting temperature in Fahrenheit and Kelvin.

Explanation / Answer

Ans 1

Part a

HCI + NaOH ---> H2O + NaCI

0.1 mole of HCI reacts with 0.1 mole of NaOH

Both reactants are in stoichiometric amount.

Moles of H2O produced = moles of HCl consumed.

= 0.10 mol

Part b

KOH + HNO3 ---> KNO3 + H2O

0.5 mol of KOH reacts with = 0.5 mol HNO3

But we have 1 mole of HNO3 which is more than required moles of HNO3

Excess reactant = HNO3

Limiting reactant = KOH

Moles of H2O produced = moles of KOH consumed

= 0.5 mol

Part C

Mass of cold water = volume x density

= 10 mL x 1g/mL = 10 g

Mass of hot water = 100 mL x 1g/mL = 100 g

Heat absorbed by cold water = Heat released by hot water

mc x Cpc x (T - T1) = mh x Cph x (T2-T)

10 x (T - 30) = 100 x (90 - T)

10T - 300 = 9000 - 100T

110T = 9300

T = 84.55 °C

Part d

Mass of cold water = volume x density

= 1gal x 3.785 L/gal x 1 g/mL

= 3.785 g

Mass of hot water = 1 L x 1000mL/L x x 1g/mL

= 1000 g

Heat absorbed by cold water = Heat released by hot water

mc x Cpc x (T - T1) = mh x Cph x (T2-T)

3.785 x (T - 0) = 1000 x (99.9 - T)

3.785 T = 99900 - 1000T

1003.785 T = 99900

T = 99.523 °C

T = 99.523 + 273.15 = 372.67 K

T = 372.67 - 457.87 = 211.136 °F

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.