Hello!
The net chemical equation between Sodium Bicarbonate and H₂SO₄ is the following one:
2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂
To calculate the amount of Sodium Bicarbonate needed to neutralize 1000 L of 0,350 M H₂SO₄ we'll need to use the following conversion factor, to go from the volume of H₂SO₄ to grams of Sodium Bicarbonate:
![1000 L H_2SO_4* \frac{0,350 moles H_2SO_4}{1 L}* \frac{2 moles NaHCO_3}{1 mol H_2SO_4}* \frac{84,007gNaHCO_3}{1molNaHCO_3} \\ \\ =58804,9 g NaHCO_3](https://tex.z-dn.net/?f=1000%20L%20H_2SO_4%2A%20%5Cfrac%7B0%2C350%20moles%20H_2SO_4%7D%7B1%20L%7D%2A%20%5Cfrac%7B2%20moles%20NaHCO_3%7D%7B1%20mol%20H_2SO_4%7D%2A%20%5Cfrac%7B84%2C007gNaHCO_3%7D%7B1molNaHCO_3%7D%20%5C%5C%20%5C%5C%20%3D58804%2C9%20g%20NaHCO_3%20)
So, to neutralize 1000 L of 0,350 moles of H₂SO₄ you'll need
58804,9 grams of NaHCO₃
Yes it correct i got it right
Explanation:
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Because they are both a reliable measurement and is both accurate and precise ... Precision, on the other hand, entails the reproducibility of an experiment.