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Rudiy27
4 years ago
14

50.0 mL of an HNo3 solution were titrated with 36.90 mL of a 0.100 M LiOH solution to reach the equivalence point. What is the m

olarity of the HNO3 solution.
Chemistry
1 answer:
sleet_krkn [62]4 years ago
3 0
Answer: 0.0738 M

Explanation:


1) Chemical equation:

<span>HNO₃ + LiOH → LiNO₃ + H₂O
</span>

2) Mole ratios:

1 mol HNO₃ neutralizes 1 mol LiOH

3) Number of moles of H⁺ from HNO₃:

M = n / V ⇒ n = M × V = 0.050 liter × M

4) Number of moles of OH⁻ from LiOH

M = n / V ⇒ n = M × V = 0.03690 liter × 0,100 M

5) Equivalence point:

number of moles of H⁺ = number of moles of OH⁻

⇒ 0.0500 × M = 0.03690 × 0.100 

⇒ M = 0.03690 × 0.100 / 0.0500 = 0.0738 M
<span />
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Answer:

             %age Yield  =  51.45 %

Solution:

Step 1: Convert Kg into g

68.5 Kg CO  =  68500 g CO

8.60 Kg H₂  =  8600 g

Step 2: Find out Limiting reactant;

The Balance Chemical Equation is as follow;

                                 CO  +  2 H₂    →    CH₃OH

According to Equation,

                   28 g (1 mol) CO reacts with  =  4 g (2 mol) of H₂

So,

                    68500 g CO will react with  =  X g of H₂

Solving for X,

                    X  =  (68500 g × 4 g) ÷ 28 g

                    X  =  9785 g of H₂

It shows 9785 g H₂ is required to react with 68500 g of CO but we are provided with 8600 g of H₂ which is less than required. Therefore, H₂ is provided in less amount hence, it is a Limiting reagent and will control the yield of products.

Step 3: Calculate Theoretical Yield

According to equation,

            4 g (2 mol) H₂ reacts to produce  =  32 g (1 mol) Methanol

So,

                          8600 g H₂ will produce  =  X g of CH₃OH

Solving for X,

                    X  =  (8600 g × 32 g) ÷ 4 g

                     X =  68800 g of CH₃OH

Step 4: Calculate %age Yield

                     %age Yield  =  Actual Yield ÷ Theoretical Yield × 100

Putting Values,

                     %age Yield  =  3.54 × 10⁴ g ÷ 68800 g × 100

                     %age Yield  =  51.45 %


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