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2 years ago

How many moles of Calcium Oxide are needed to produce 4 moles of Calcium Hydroxide?

Chemistry

1 answer:

Tpy6a [65]2 years ago

5 0

Taking into account the reaction stoichiometry, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CaO + H₂O → Ca(OH)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CaO: 1 mole
H₂O: 1 mole
Ca(OH)₂: 1 mole

<h3>Moles of CaO required</h3>

The following rule of three can be applied: If by stoichiometric reaction 1 mole of Ca(OH)₂ is produced by 1 mole of CaO, 2 moles of Ca(OH)₂ are produced by how many moles of CaO?

$moles of CaO=\frac{2 moles of Ca(OH)_{2}x1 mol of CaO }{1 mole of Ca(OH)_{2}}$

moles of CaO= 2 moles

Finally, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

Learn more about the reaction stoichiometry:

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What is the number of moles in 432g Ba(NO3)2

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First there is a need to calculate the molar mass of Ba(NO₃)₂:

137.3 + 2 (14.0) + 6 (16) = 261.3 grams/mole

The molar mass, denoted by M in chemistry refers to a physical characteristic illustrated as the mass of a given component divided by the amount of the component. The molar masses are always denoted in grams/mole.

After finding the molar mass, the number of moles can be identified as:

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12oz of water initially at 75oF is mixed with 20oz of water intiially at 140oF. What is the final temperature?

Kaylis [27]

Answer:

$115.625^{\circ}\text{F}$

Explanation:

$m_1$ = First mass of water = 12 oz

$m_2$ = Second mass of water = 20 oz

$\Delta T_1$ = Temperature difference of the solution with respect to the first mass of water = $(T-75)^{\circ}\text{F}$

$\Delta T_2$ = Temperature difference of the solution with respect to the second mass of water = $(T-75)^{\circ}\text{F}$

c = Specific heat of water

As heat gain and loss in the system is equal we have

$m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}$

The final temperature of the solution is $115.625^{\circ}\text{F}$ .

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