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.

Answer 1

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