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Balancing Chemical Equations Balance the equation Na(s) +

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 1PE Chapter 3.2SE

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 1PE

Balancing Chemical Equations Balance the equation Na(s) + H20(/)—? NaOH(aq) + H2 (g) The unbalanced equation for the reaction between methane and bromine is Once this equation is balanced what is the value of the coefficient in front of bromine Br2? (a)1, (b) 2, (c) 3, (d) 4, (e) 6.

Step-by-Step Solution:
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Answer Begin by counting each kind of atom on the two sides of the arrow. There are one Na, one O, and two H on the left side, and one Na, one O, and three H on the right. The Na and O atoms are balanced, but the number of H atoms is not. To increase the number of H atoms on the left, let’s try placing the coefficient 2 in front of H2O: Na(s)+2H2O(l) NaOH(aq)+H2(g). Although beginning this way does not balance H, it does increase the number of reactant H atoms, which we need to do. (Also, adding the coefficient 2 on H2O unbalanced O, but we will take care of that after we balance H.) Now that we have 2 H2O on the left, we balance H by putting the coefficient 2 in front of NaOH: Na (aq)+2H2O2NaOH+H2 Balancing H in this way brings O into balance, but now Na is unbalanced, with one Na on the left and two on the right. To rebalance Na, we put the coefficient 2 in front of the reactant: 2Na+2H2O 2NaOH + H2 We now have two Na atoms, four H atoms, and two O atoms on each side. The equation is balanced. Comment Notice that we moved back and forth, placing a coefficient in front of H2O, then NaOH, and finally Na. In balancing equations, we often find ourselves following this pattern of moving back and forth from one side of the arrow to the other, placing coefficients first in front of a formula on one side and then in front of a formula on the other side until the equation is balanced. You can always tell if you have balanced your equation correctly by checking that the number of atoms of each element is the same on the two sides of the arrow, and that you’ve chosen the smallest set of coefficients that balances the equation. In the next question we have to balance the given reaction . CH4+Br2CBr4+HBr Here in the given reaction we have One carbon atom ,4 hydrogen atoms ,and 2 bromine atoms on the reactant side and one carbon atom ,5 bromine atoms ,and 1 hydrogen atom on the right hand side of the reaction we have to balance the reaction starting from hydrogen atom. There are four hydrogen atoms on the reactant side but only one hydrogen atom on product side. So, multiply the HBr on product side by the coefficient 4. Then the equation becomes as follows: CH4(g)+Br2(l)CBr4(s)+4HBr(g) Now, carbon and hydrogen atoms are balanced. Balance the bromine atoms in the equation. There are two bromine atoms on the reactant side but eight atoms of bromine are there on product side. Thus, multiply bromine molecule on reactant side by coefficient 4. Then the equation becomes a balanced chemical equation which is represented as follows: CH4+4Br2(l)CBr4(s)+4HBr(g) The coefficient in front of bromine, Br2 is 4.

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Chapter 3.2SE, Problem 1PE is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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