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ruslelena [56]
3 years ago
15

50.0 mL of a solution of HCl is combined with 100.0 mL of 1.05M NaOH in a calorimeter. The reaction mixture is initially at 22.4

degree C, and the final temperature after reaction is 30.2 degree C. What is the molarity of the HCl solution? You may assume that there is an excess of base (so that all the HCl has reacted), that the specific heat of the reaction mixture is 0.96 cal/g degree C, and that the density of the reaction mixture is 1.02 g/mL. The heat of neutralization of HC1 and NaOH is 13.6 kcal/mole.
Chemistry
1 answer:
VikaD [51]3 years ago
3 0

Answer:

2.1 M is the molarity of the HCl solution.

Explanation:

HCl+NaOH\rightarrow H_2O+NaCl

Molarity of HCl solution = M_1=?

Volume of HCl solution = V_1=50.0mL

Ionizable hydrogen ions in HCl = n_1=1

Molarity of NaOH solution = M_2=1.05 M

Volume of NaOH solution = V_2=100.0 mL

Ionizable hydroxide ions in NaOH = n_2=1

n_1M_1V_1=n_2M_2V_2 (neutralization )

M_1=\frac{M_2V_2}{V_1}=\frac{1.05M\times 100.0 mL}{50.0 mL}

M_1=2.1 M

2.1 M is the molarity of the HCl solution.

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When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a
AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0
3 years ago
What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0
2 years ago
Complete the following single replacement reaction. If they don’t react, just write “NR”
stiv31 [10]

We have to complete all the given reactions.

1. Fe(s) + CuCl₂ → Cu + FeCl₂

2. Cu(s) + FeCl₂(aq)  → NR (no reaction takes place)

3. K(s) + NiBr2(aq) → NR (no reaction takes place)

4. Ni(s) + KBr(aq) → K + NiBr₂

5. Zn(s) + Ca(NO₃)₂(aq) → NR (no reaction)

6. Ca(s) + Zn(NO₃)₂(aq) → Zn(s) + Ca(NO₃)₂(aq)

4 0
3 years ago
8. A sample of sulfur has a mass of 223 g. How many moles of sulfur are in the sample?
Rama09 [41]

Answer:

7 moles

Explanation:

1mole of sulfur=32

x=223

=6.9

=7moles

8 0
3 years ago
What claim is being made?
dangina [55]
The correct answer is c
7 0
3 years ago
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