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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 53e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 53e

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Calculate the molarity of each solution.a. 3.25 mol of ISBN: 9780321809247 1

Solution for problem 53E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Problem 53E

Calculate the molarity of each solution.

a. 3.25 mol of LiCl in 2.78 L solution

b. 28.33 g C6H12O6 in 1.28 L of solution

c. 32.4 mg NaCl in 122.4 mL of solution

Step-by-Step Solution:
Step 1 of 3

Solution 53E: Here, we are going to write the molecular and net ionic equations for the given set of reactions and also identify the gas formed in each case. Step1: Net ionic equation shows only those chemical species which are taking part in a chemical reaction. while writing the ionic equation, we write out all the soluble compounds as ions and eliminate ions common to both the reactants and products. This will give us the net ionic equation. Step2: a) Manganese reacts with dilute sulfuric acid to form manganese sulfate and hydrogen gas. The balanced molecular equation is given below: Mn(s) + dil H SO (aq) ---------> MnSO (aq) + H (g) 2 4 4 2 From the rules of solubility, we can say that, H SO and MnSO are w2er 4 uble 4 compounds and will dissociate into ions in aqueous solution. So, the balanced ionic equation for the above reaction is: Mn + 2H + SO 4----------> Mn + SO+ 42-+ H 2 Eliminating the spectator ions on both sides of the equation, we get the net ionic equation as follows: Mn + 2H ---------> Mn + H 2+ 2 b) Chromium reacts with hydrobromic acid to form chromium bromide and hydrogen gas. The balanced molecular equation is given below: 2Cr(s) + 6HBr(aq) ---------> 2CrBr (aq) + 3H (g) 3 2 From the rules of solubility, we can say that, HBrand CrBr are water soluble 3 compounds and will dissociate into ions in aqueous solution. So, the balanced ionic equation for the above reaction is: + - 3+ - 2Cr + 6H + 6Br ---------> 2Cr + 6Br + 3H 2 Eliminating the spectator ions on both sides of the equation, we get the net ionic equation as follows: + 3+ 2Cr + 6H ---------> 2Cr + 3H 2 c) Tin reacts with hydrochloric acid to form stannous chloride and hydrogen gas. The balanced molecular equation is given below: Sn(s) + 2HCl(aq) ---------> SnCl (aq) + H (g)2 2 From the rules of solubility, we can say that, HCland SnCl are water soluble 2 compounds and will dissociate into ions in aqueous solution. So, the balanced ionic equation for the above reaction is: + - 2+ - Sn + 2H + 2Cl ---------> Sn + 2Cl + H 2 Eliminating the spectator ions on both sides of the equation, we get the net ionic equation as follows: Sn + 2H ---------> Sn + H 2+ 2 d) Aluminium reacts with formic acid to form aluminium formate and hydrogen gas. The balanced molecular equation is given below: 2Al(s) + 6HCOOH(aq) ---------> 2Al(HCOO) (aq) + 3H (g) 3 2 From the rules of solubility, we can say that, Al(HCOO) is water soluble 3 compounds and will dissociate into ions in aqueous solution. So, the balanced ionic equation for the above reaction is: 3+ - 2Al(s) + 6HCOOH(aq) ---------> 2Al + 6HCOO(aq) + 3H (g) 2 Since, no spectator ions are present, so the above derived equation is the net ionic equation.

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Step 3 of 3

ISBN: 9780321809247

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