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ella [17]
3 years ago
15

Calculate how many grams of each product are formed when 4.05g of h2o is used. 2h2o=2h2+o2

Chemistry
1 answer:
NISA [10]3 years ago
3 0

Answer:

Explanation:

Given data:

Mass of water used = 4.05 g

Mass of each product produced = ?

Solution:

Chemical equation:

2H₂O   →  2H₂ + O₂

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 4.05 g/ 18 g/mol

Number of moles = 0.225 mol

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

                             H₂O            :              H₂

                               2               :               2

                           0.225           :           0.225

                           

                           H₂O             :              O₂

                             2                :               1

                           0.225           :            1/2×0.225 = 0.113 mol

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 0.225 × 2 g/mol

Mass = 0.45 g

Mass of oxygen:

Mass = number of moles ×  molar mass

Mass = 0.113 mol × 32 g/mol

Mass = 3.616 g

                     

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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

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K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

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1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

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This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

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0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

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