Balance the following chemical equation using the steps in balancingequations. 1. Fe + Na Br→ Fe Brz + Na2. H2 + O2→ H2 O3. Hg O--→ Hg + O2 BALANCING EQUATIONS:Steps in balancing equations:Take note of the elements present in thereactant and product sideExample:Na CI + Ba OВа ClaCount the number of atom/s of eachelement present in the reactant andproduct sideProductsReactantsNaCINa→ Naz o +ВаВаCINa ---- 1 atomNa--- 2 atomsCI-1 atom1 atomВаCI -------- 2 atomsВа--- 1 atom1 atom------- 1 atomApply the Law of Conservation ofMass to get the same number of atomsof every element on each side of theequation.1. Ba and O appear to balanced, proceed to balancedNa & CI2. To balanced Na & CI, find their LCM (Least CommonMultiple) which is 2,13. The LCM of 2 & 1 is 2To check:Na-- 1 atom x 2 = 2 atoms Na ---2 atoms x 1=2 atomsCI -- 1 atom x 2 = 2 atoms o -- 1 atomBa-- 1 atomO --- 1 atomBalance the chemical equations byplacing the appropriate coefficientsbefore the symbol or formulaВа -- 1 atomDo not change the subscripts of theformulaCI--- 2 atoms x 1=2 atomsSupply the numbers 2 as coefficients in the equation2 Na CI + Ba O -------→ 1 Na, O + Ba Cl,The chemical equation is nowbalanced.It follows the Law of2 Na CI + Ba o-3 Naz O + Bа ClzConservation of Mass.

Answered: 1. Fe + Na Br - Fe Brs + Na 2. Hz + O2…

Science

Chemistry

1. Fe + Na Br - Fe Brs + Na 2. Hz + O2 - H2 O 3. Hg O ------→ Hg + 02

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter13: Chemical Equilibrium

Section: Chapter Questions

Problem 128IP: The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered...

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al Calculate ho AH far the fellowreg reaction usivg Heao Law: i) CS2+2H2O PCO2 + 2H2S Hre fellaweng HiS+ Ź O2 tHi0+502 AH=-563KJ info: C5z +302+ O2 + 2502 Ato-l015KJ CHut 3C124 CH Cl 3 f 3HC) Givan the Fallavine info: Ct 2H2 e CHy AH: – 14.8KJ Ct Ź Clz t Ź Hz e cH as oH=-132.1 KJ AH= -920KJ 3
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