All categories

2 years ago

If you react 156.0 g of calcium chloride with an excess NaOH, how much sodium chloride should you get?

Chemistry

1 answer:

laiz [17]2 years ago

7 0

Answer:

164.3g of NaCl

Explanation:

Based on the chemical equation:

CaCl2 + 2NaOH → 2NaCl + Ca(OH)2

where 1 mole of CaCl2 reacts with 2 moles of NaOH

To solve this question we must convert the mass of CaCl2 to moles. Using the chemical equation we can find the moles of NaCl and its mass:

Moles CaCl2 -Molar mass: 110.98g/mol-

156.0g CaCl₂ * (1mol / 110.98g) = 1.4057 moles CaCl2

Moles NaCl:

1.4057 moles CaCl2 * (2mol NaCl / 1mol CaCl2) = 2.811 moles NaCl

Mass NaCl -Molar mass: 58.44g/mol-

2.811 moles NaCl * (58.44g / mol) = 164.3g of NaCl

You might be interested in

Ehem I have thirty levels in minecraft and a full enchantment room set up what should I put it on? My diamond axe (which is my w

loris [4]

Probably the diamond pickaxe

7 0

3 years ago

Read 2 more answers

I. Choose the letter of the correct answer. Write your answer on a separate sheet of paper.

omeli [17]

Answer:

,,,

Explanation:

6 0

2 years ago

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

3 years ago

What is structure 1, and how does it help sustain life

Bezzdna [24]

Answer:Chloroplast

Explanation:

5 0

2 years ago

Read 2 more answers

10 examples of consumers

ivann1987 [24]

Squirrel, lion, turtle , crab, jelly fish, dolphin, fish, cow, rooster, human

5 0

3 years ago

Read 2 more answers