How many grams of sodium carbonate (Na2CO3) are needed to produce 6.78

Chemistry

1 answer:

olya-2409 [2.1K]3 years ago

7 0

Answer:

359.3g of Na₂CO₃ are required

Explanation:

When sodium carbonate, Na₂CO₃ reacts with HCl, NaCl and H₂O are produced as follows:

Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

<em>Where 1 mole of sodium carbonate produce 2 moles of NaCl</em>

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To solve this question, we must find the moles of sodium carbonate required to produce the 6.78 moles of NaCl. With this mass and the molar mass we can find the mass of sodium carbonate as follows:

<em>Moles Na₂CO₃:</em>

6.78 moles NaCl * (1mol Na₂CO₃ / 2mol NaCl) = 3.39 moles Na₂CO₃

<em>Mass Na₂CO₃ -Molar mass: 105.99g/mol:</em>

3.39 moles Na₂CO₃ * (105.99g / mol) =

<h3>359.3g of Na₂CO₃ are required</h3>

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For the reaction, 2cr2+ + cl2(g) ---> 2cr3+ + 2cl- e cell (standard conditions) = 1.78v calculate ecell (standard conditions)

PolarNik [594]

Answer:

-1.78 V

Explanation:

There are several rules required to calculate the cell potential:

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That said, notice that the initial reaction with respect to the final reaction is:

reversed: chromium(III) cation and chloride anion become our reactants as opposed to the products in the initial reaction, so we change the sign of the cell potential to a negative value of -1.78 V;
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