Would someone mine answering all of theses questions for me its extremely import

Question

Would someone mine answering all of theses questions for me its extremely important omework- Solution Stoichiometry hese problems are similar to what you have done in lab, but not identical. If you need more formation, refer to your General Chemistry textbook for assistance. o receive full credit, you must SHOW ALL YOUR WORKII Name If 16.97 g of NaBr is dissolved in enough water to make 160. ml of solution, what is the molar concentration of NaBr? 1. b. How many mililiters of 0.10 M NaBr would you need to supply 6.31 g of NaBr? 2. For each of the following preparations, tell how many grams of solute would be necessary for its preparation. a. 0.90 L of 0.15 M BaCl, b. 50 ml of 0.018M KMnO 3. What would be the molar concentrations of the solute in each of the following solutions? a. 0.so L containing 38.6 g of NaCIO b. 0.25 L containing 41.S g of CHO 4. If you require 500. ml of 0.25 M NaOH, how many milliliters of 1.39 M NaOH must you dilute? 123

Explanation / Answer

1a) mol NaBr= mass/molecular wheight= 16.97g/103g/mol= 0.165 mol

Molarity= mol/Volume (L)= 0.165mol/0.16L = 1.03 M

1b) mol NaBr= 6.31g/103g/mol= 0.061 mol

V= 0.061 mol/0.1M = 0.61L -----> 610 mL

2a) mol BaCl2= 0.15M x 0.9L = 0.135 mol ----> mass= 0.135mol x 207g/mol= 27.945g of solute

2b) mol KMnO4= 0.018M x 0.05L= 9x10-4 mol ----> mass= 9x10-4 mol x 158g/mol= 0.1422g of solute

3a) mol NaClO4= 38.6g/122g/mol= 0.316 mol ----> Molarity= 0.316mol/0.5L= 0.632M

3b) mol C4H8O= 41.5g/72g/mol= 0.576M

4) Molarity1 x Volume 1= Molarity 2 x Volume 2

On this equation you just have to put to one side the information of the solution you want to prepare, in this case 500mL of NaOH 0.25M, and at the other side the information that you have from the concentrated solution that you are going to dilute. Let´s see:

1.39M x V= 0.25Mx 0.5L

V= 0.09 L = 9 mL

5) First of all let´s balance the equation:

Mg + 2HCl -----> MgCl2 + H2

2 mol HCl ------------ 1 mol H2

x= 0.0242 mol HCl------------ 0.0121 mol

V of HCl= 0.0242 mol /2M= 0.0121 L ----> 12.1 mL

6) Na2CO3 + 2HCl -----> 2NaCl + CO2 + H2O

mol Na2CO3= 3.8M x 0.1L= 0.38 mol

1 mol Na2CO3 -------- 2 mol HCl

0.38 mol Na2CO3 --------- x=0.76 mol HCl ------> V HCl= 0.76mol/0.92M= 0.826 L ----> 826 mL

7) mol HCl= 0.75M x 0.045L= 0.03375 mol

1 mol Na2CO3 -------- 2 mol HCl

x= 0.016875 mol -------- 0.03375 mol HCl

mass Na2CO3= 0.016875 mol/106g/mol= 1.59x10-4g

1 mol Na2CO3 -------- 2 mol NaCl

0.016875 mol ---------- x= 0.03375 mol NaCl -----> mass= 0.03375mol x 58g/mol= 1.96 g

8a) mol NaOH=0.953M x 0.04133L= 0.039 mol

1 mol of acetic acid reacts with 1 mol of NaOH so, 0.039 mol is the amount of acetic acid on the sample.

mass acetic acid= 0.039 mol x 60g/mol= 2.34g

8b) M= 0.039mol/0.025L= 1.56M

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