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morpeh [17]
3 years ago
15

Grams of solute needed

Chemistry
1 answer:
scZoUnD [109]3 years ago
5 0

Answer:

grams of solute = 31.5 g

Explanation:

Given data:

Grams of solute = ?

%m/v  = 3%

Volume = 1050 mL

Solution:

grams of solute = 3 g of NH₄Cl / 100 mL × 1050 mL

grams of solute = 0.03 g × 1050  

grams of solute = 31.5 g

So 31.5 g are needed to to make 3% m/v solution.

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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:

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Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

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4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

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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

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