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jarptica [38.1K]
3 years ago
10

According to the equation below, how many moles of Ca(OH)2 are required to react with 1.36 mol H3PO4 to produce Ca3(PO4)2? 3Ca(O

H)2+2H3PO4⟶Ca3(PO4)2+6H2O
Chemistry
1 answer:
allsm [11]3 years ago
3 0

<u>Answer:</u> The amount of calcium hydroxide needed to react is 2.04 moles

<u>Explanation:</u>

We are given:

Moles of phosphoric acid = 1.36 moles

For the given chemical equation:

3Ca(OH)_2+2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O

By Stoichiometry of the reaction:

2 moles of phosphoric acid reacts with 3 moles of calcium hydroxide

So, 1.36 moles of phosphoric acid will react with = \frac{3}{2}\times 1.36=2.04mol of calcium hydroxide

Hence, the amount of calcium hydroxide needed to react is 2.04 moles

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On a hot summer day you and some friends decide you want to cool down your pool. Determine the mass of ice you would need to add
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Answer: The mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is 16.14 \times 10^{4} kg.

Explanation:

We know that relation between heat energy and specific heat is as follows.

                 q = m \times S \times \Delta T

As density of water is 1 kg/L and volume is given as 400,000 L. Therefore, mass of water is as follows.

          Mass of water = Volume × Density

                                  = 400,000 L \times 1 kg/L

                                  = 400,000 kg

or,                              = 400,000 \times 10^{3} g    (as 1 kg = 1000 g)

Specific heat of water is 4.2 J/gm K. Therefore, change in temperature is as follows.

         \Delta T = 305 K - 273 K

                    = 32 K

Now, putting the given values into the above formula and calculate the heat energy as follows.

            q = m \times S \times \Delta T

                = 400,000 \times 10^{3} \times 4.2 \times 32 K

                = 5376 \times 10^{7} J

or,            = 5376 \times 10^{4} kJ

According to the enthalpy of melting of ice 333 kJ/Kg of energy absorbed by by 1 kg of ice. Hence, mass required to absorb energy of 5376 \times 10^{4} kJ  is calculated as follows.

            Mass = \frac{5376 \times 10^{4} kJ}{333 kJ/Kg} \times 1 kg

                      = 16.14 \times 10^{4} kg

Thus, we can conclude that the mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is 16.14 \times 10^{4} kg.

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