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Write a balanced equation for combustion of C8H7NO2SBrCIin a C,H,N,Selemental analyzer.

Short Answer

Expert verified

The balanced equation for combustion of is

C8H7NO2SBrCI+914O2→8CO2+52H2O+12N2+SO2+HBr+HCl.

Step by step solution

01

Defining the balanced equation.

A balanced equation is a chemical reaction equation in which the total charge and the number of atoms for each element in the reaction are the same for both the reactants and the products. In other words, the mass and charge on both sides of the reaction are balanced.

02

Writing a balanced equation for combustion of  C8H7NO2SBrCI.

  • Combustion, also known as burning, is a high-temperature exothermic redox chemical reaction that occurs between a fuel (the reductant) and an oxidant, usually atmospheric oxygen, to produce oxidised, often gaseous products in a mixture known as smoke.
  • The combustion reaction occurs when fuel and oxygen react, resulting in fire, heat, and light.
  • Combustion occurs when gasoline, which is typically a fossil fuel, reacts with the oxygen in the air to produce heat.
  • The heat generated by the combustion of fossil fuels is used to power machinery like boilers, furnaces, ovens, and engines.

The balanced equation for combustion ofC8H7NO2SBrCIis

C8H7NO2SBrCI+914O2→8CO2+52H2O+12N2+SO2+HBr+HCl.

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Most popular questions from this chapter

Write a balanced equation for the combustion of benzoic acid,C6H5CO2H, to giveCO2 and H2O. How many milligrams of and of will be produced by the combustion of 4.635mgof benzoic acid?

Man in the vat problem.15 Long ago, a workman at a dye factory fell into a vat containing hot, concentrated sulfuric and nitric acids. He dissolved completely! Because nobody witnessed the accident, it was necessary to prove that he fell in so that the man’s wife could collect his insurance money. The man weighed 70 kg, and a human body contains |6.3 parts per thousand (mg/g) phosphorus. The acid in the vat was analyzed for phosphorus to see whether it contained a dissolved human.

(a) The vat contained8.00×103Lof liquid, and a 100.0-mL sample was analyzed. If the man did fall into the vat, what is the expected quantity of phosphorus in 100.0 mL?

(b) The 100.0-mL sample was treated with a molybdate reagent that precipitated ammonium phosphomolybdate,(NH4)3[P(Mo12O40)]12H2OThis substance was dried at110°Cto remove waters of hydration and heated to400°Cuntil it reached the constant compositionP2O5×24MoO3, which weighed 0.371 8 g. When a fresh mixture of the same acids (not from the vat) was treated in the same manner, 0.033 1 g ofP2O5×24MoO3(FM3596.46)was produced. This blank determination gives the amount of phosphorus in the starting reagents. TheP2O5×24MoO3that could have come from the dissolved man is therefore0.3718-0.0331=0.3387g.How much phosphorus was present in the 100.0-mL sample? Is this quantity consistent with a dissolved man?

What measures can be taken to decrease the relative supersaturation during a precipitation?

Combustion analysis of a compound known to contain justC,H,N, and Odemonstrated that it is 46.21wt%C,9.02wt% H, 13.74wt N and, by difference, 100 - 46.21 - 9.02 -13.74 = 31.03% O. This means that of unknown would contain46.21gof C,9.02g ofH, and so on. Find the atomic ratio C : H : N : Oand express it as the lowest reasonable integer ratio.

Combustion analysis of an organic compound gave the composition 71.17±0.41wt%C,6.76±0.12wt%H, and10.34±0.08wt%N . Find the coefficients and and their uncertainties and in the formula C8H±±xNn±y.

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