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Lelechka [254]
3 years ago
5

Barium nitrate and sodium sulfate solutions can be used to precipitate barium sulfate. How many grams

Chemistry
1 answer:
-Dominant- [34]3 years ago
7 0

Answer:

40.8g of sodium sulfate must be added

Explanation:

The reaction of barium nitrate, Ba(NO₃)₂ with sodium sulfate, Na₂SO₄ is:

Ba(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + BaSO₄(s)

That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:

<em>Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:</em>

75g * (1mol / 261.3g) = 0.287 moles = Moles Na₂SO₄

<em>Mass Na₂SO₄ -Molar mass: 142.04g/mol-:</em>

0.287 moles * (142.04g / mol) =

<h3>40.8g of sodium sulfate must be added</h3>
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Determine the oxidation state of the metal ion in [co(nh3)5br]2+. express your answer as an integer.
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                                           [Co(NH₃)₅Br]²⁺

Ligands and charges on them,

                             5 × NH₃  =  5 × 0  =  0

                             1 × Br⁻¹  =  1 × -1  =  -1

Charge on sphere  =  +2

So, putting values in equation,

                                       Co + (0)₅ - 1  =  +2

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                                       Co - 1  =  +2

                                       Co  =  +2 + 1

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What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
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Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

3 0
3 years ago
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