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polet [3.4K]
2 years ago
13

Aqueous solutions of barium nitrate and potassium phosphate are mixed. What is the precipitate and how many molecules are formed

?
Chemistry
2 answers:
kykrilka [37]2 years ago
8 0

Aqueous solutions of barium nitrate and potassium phosphate are mixed. What is the precipitate and how many molecules are formed?

Barium nitrate has a chemical symbol of Ba(NO3)2 and potassium phosphate has a chemical symbol K2PO4. The reaction between these two is a double replacement reaction yielding barium phosphate and potassium nitrate.

The chemical equation representing the reaction is,

            Ba(NO3)2 + K2PO4 à KNO3 + BaPO4

Anna007 [38]2 years ago
6 0

Answer: The precipitate is Ba_3(PO_4)_2 and 1 molecule of Barium phosphate and 6 molecules of Potassium nitrate is formed.

Explanation: We are given Barium nitrate and Potassium phosphate , that will lead to the formation of Barium phosphate and Potassium nitrate.

As we know that these solutions are present in water, therefore we will obtain Barium phosphate as a precipitate because Barium phosphate is insoluble in water and Potassium nitrate is soluble in water. Thus the reaction is

Ba(NO_3)_2(aq.)+K_3PO_4(aq.)\rightarrow Ba_3(PO_4)_2(ppt.)+KNO_3(aq.)

To know how many molecules are formed at the end, we need to balance the above equation.

Balancing the equation, we get:

3Ba(NO_3)_2(aq.)+2K_3PO_4(aq.)\rightarrow Ba_3(PO_4)_2(ppt.)+6KNO_3(aq.)

Thus, we get 1 molecule of Barium phosphate as a precipitate and 6 molecules of Potassium nitrate which is easily soluble in water.


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2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
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Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

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PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

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<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

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The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2
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Answer:

                    No the substance is not water.

Explanation:

                   The balance chemical equation for the decomposition of water is as follow;

                                           2 H₂O = 2 H₂ + O₂

Step 1: <u>Calculate moles of H₂O;</u>

               Moles  =  Mass / M.Mass

               Moles  =  5.0 g / 18.01 g/mol

               Moles  =  0.277 moles of H₂O

Step 2: <u>Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:</u>

According to equation,

                        2 moles of H₂O produced  =  1 mole of O₂

So,

                  0.277 moles of H₂O will produce  =  X moles of O₂

Solving for X,

                     X =  0.277 mol × 1 mol / 2 mol

                     X =  0.138 moles of O₂

Also,

According to equation,

                        2 moles of H₂O produced  =  2 mole of H₂

So,

                  0.277 moles of H₂O will produce  =  X moles of H₂

Solving for X,

                     X =  0.277 mol × 2 mol / 2 mol

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Step 3: <u>Calculate Mass of O₂ and H₂ as;</u>

For O₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.138 mol × 31.99 g/mol

                 Mass  =  4.44 g of O₂

For H₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.227 mol × 2.01 g/mol

                 Mass  =  0.559 g of H₂

Conclusion:

                   From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

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