A 25.00-mL sample of an H2SO4 solution of unknown concentration is titrated with a 0.1322 M KOH solution. A volume of 41.22 mL o
1 answer:
<u>Answer:</u> The concentration of comes out to be 0.109 M.
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given y neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
Hence, the concentration of comes out to be 0.109 M
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Answer:
5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃
6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693. grams of N₂
Explanation:
5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃) is presented as follows;
(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O
One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂
The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;
The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g
6. The chemical equation for the reaction is presented as follows;
N₂ + 3H₂ → 2NH₃
Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)
The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;
The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693. grams