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Solved and balanced chemical equation 3 As2S3 + 28 HNO3 + 4 H2O → 6 H3AsO4 + 9 H2SO4 + 28 NO with completed products. Application for completing products and balancing equations.
- As2S3 + 28 HNO3 → 2 H3AsO4 + 3 H2SO4 + 28 NO2 + 8 H2O - Balanced ...
Solved and balanced chemical equation As2S3 + 28 HNO3 → 2...
- As2S3 + 28 HNO3 → 2 H3AsO4 + 3 H2SO4 + 28 NO2 + 8 H2O - Balanced ...
30 Ιουν 2021 · Basic concepts of chemistry and chemical calculation. Book exercise. As2S3 + HNO3 + H2O → H3AsO4+H2SO4+NO ...more.
To be balanced, every element in As2S3 + H2O + HNO3 = NO + H3AsO4 + H2SO4 must have the same number of atoms on each side of the equation. When using the inspection method (also known as the trial-and-error method), this principle is used to balance one element at a time until both sides are equal and the chemical equation is balanced.
a As 2 S 3 + b H 2 O + c HNO 3 = d H 3 AsO 4 + e NO + f H 2 SO 4 Now we write down algebraic equations to balance of each atom: As: a * 2 = d * 1 S: a * 3 = f * 1 H: b * 2 + c * 1 = d * 3 + f * 2 O: b * 1 + c * 3 = d * 4 + e * 1 + f * 4 N: c * 1 = e * 1 Now we assign a=1 and solve the system of linear algebra equations: a * 2 = d a * 3 = f b ...
The stoichiometric mole ratio of As2S3 and HNO3 for a maximum theoretical yield is 3:22, which will yield H3AsO4, SO2, NO and H2O in a ratio of 3:22:6:9.
The stoichiometric mole ratio of As2S3, HNO3 and H2O for a maximum theoretical yield is 3:10:4, which will yield NO, H3AsO4 and S in a ratio of 3:10:4.
Solved and balanced chemical equation As2S3 + 28 HNO3 → 2 H3AsO4 + 3 H2SO4 + 28 NO2 + 8 H2O with completed products. Application for completing products and balancing equations.
