How many grams of na2co3 would be needed to produce 1000g of nahco3
1 answer:
Answer:
630.95 grams of Na₂CO₃ would be needed to produce 1000g of NaHCO₃
Explanation:
The balanced reaction is:
Na₂CO₃ + CO₂+ H₂O → 2 NaHCO₃
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- Na₂CO₃: 1 moles
- CO₂: 1 mole
- H₂O: 1 mole
- NaHCO₃: 2 moles
Being the molar mass:
- Na₂CO₃: 106 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
- NaHCO₃: 84 g/mole
Then by stoichiometry the following quantities of mass participate in the reaction:
- Na₂CO₃: 1 mole* 106 g/mole= 106 g
- CO₂: 1 mole* 44 g/mole= 44 g
- H₂O: 1 mole* 18 g/mole= 18 g
- NaHCO₃: 2 moles* 84 g/mole= 168 g
You can apply the following rule of three: if 106 grams of Na₂CO₃ are needed to produce 168 grams of NaHCO₃, how much mass of Na₂CO₃ is necessary to produce 1000 grams of NaHCO₃?
mass of Na₂CO₃= 630.95 grams
<u><em>630.95 grams of Na₂CO₃ would be needed to produce 1000g of NaHCO₃</em></u>
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Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
- V₂ = 295 mL
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L