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lions [1.4K]
3 years ago
12

A chemistry student weighs out of hypochlorous acid into a volumetric flask and dilutes to the mark with distilled water. He pla

ns to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
3 0

Answer:

<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>

A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.

Explanation:

1 mole HClO = 74.44g

0.0941g = \frac{0.0941}{74.44} = 0.00126 moles

Concentration = no. of moles/volume in L

Hence, Concentration of HClO = 0.00126/ 0.250L

= 0.005M.

C1V1 =C2V2

0.005 × 250 mL = 0.2 × V2

<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>

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What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH
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1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
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<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover. 

                                  CO       +         2H2   ------->      CH3OH  
                                 1 mol              2 mol
given                        2.66 mol          6 mol (excess)

How much
we need  CO?           3 mol              6 mol

We see that H2 will be leftover, because for 6 moles H2  we need 3 moles CO, but we have only 2.66 mol  CO.
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                                 1 mol                                        1 mol
                                2.66 mol                                    2.66 mol

4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 +  4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH =  85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
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