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ladessa [460]
3 years ago
9

How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?

Chemistry
1 answer:
Andre45 [30]3 years ago
6 0

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

     S1=50%

      V2=250ml

      S2= 20%

We all know that

                     V1S1=V2S2

                     ∴V1=  V2×S2÷S1

                     ∴V1=  V2S2×1/S1

                      ∴V1= 250×20÷50

                       ∴V1= 100ml

 

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Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl4
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Answer:

a) 226.6 grams of Cl₂

b) 19.2 grams of C

c) 303.2 grams of TiCl₄ and 70.4 grams of CO₂

Explanation:

The balanced chemical reaction is the following:

TiO₂(s) + C(s) + 2 Cl₂(g) → TiCl₄(s) + CO₂(g)

(a) What mass of Cl₂ gas is needed to react with 1.60 mol TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 2 moles of Cl₂. So, the stoichiometric ratio is 2 mol Cl₂/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of Cl₂ we need:

1.60 mol TiO₂ x 2 mol Cl₂/1 mol TiO₂ = 3.2 mol Cl₂

Now, we convert from moles to mass by using the molecular weight (MW) of Cl₂:

MW(Cl₂) = 35.4 g/mol x 2 = 70.8 g/mol

mass of Cl₂= 3.2 mol x 70.8 g/mol = 226.6 g

<em>Therefore, 226.6 grams of Cl₂ are needed to react with 1.6 mol of TiO₂. </em>

(b) What mass of C is needed to react with 1.60 mol of TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 1 moles of C(s). So, the stoichiometric ratio is 1 mol C/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of C(s) we need:

1.60 mol TiO₂ x 1 mol C(s)/1 mol TiO₂ = 1.60 mol C(s)

So, we convert the moles of C(s) to grams as follows:

MW(C) = 12 g/mol

1.60 mol x 12 g/mol = 19.2 g C(s)

<em>Therefore, a mass of 19.2 grams of C is needed to react with 1.60 mol of TiO₂. </em>

(c) What is the mass of all the products formed by reaction with 1.60 mol of TiO₂?

From the chemical equation, we can notice that 1 mol of TiO₂ produces 1 mol of TiCl₄ and 1 mol of CO₂. So, from 1.60 moles of TiO₂, 1 mol of each product will be produced:

1 mol TiO₂/1 mol TiCl₄ ⇒ 1.60 mol TiO₂/1.60 mol TiCl₄

1 mol TiO₂/1 mol CO₂ ⇒ 1.60 mol TiO₂/1.60 mol CO₂

Finally, we convert the moles to grams by using the molecular weight of each compound:

MW(TiCl₄) = 47.9 g/mol Ti + (35.4 g/mol x 4 Cl) = 189.5 g/mol

1.60 mol x 189.5 g/mol = 303.2 g

MW(CO₂) = 12 g/mol C + (16 g/mol x 2 O) = 44 g/mol

1.60 mol x 44 g/mol = 70.4 g

<em>Therefore, from the reaction of 1.60 mol of TiO₂ are formed 303.2 grams of TiCl₄ and 70.4 grams of CO₂.</em>

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