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2 years ago

Please help...awarding brainliest if correct and lots of points.

Chemistry

1 answer:

abruzzese [7]2 years ago

8 0

Silver chloride produced : = 46.149 g

Limiting reagent : CuCl2

Excess remains := 3.74 g

<h3>Further explanation</h3>

Reaction

silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate

Required

silver chloride produced

limiting reagent

excess remains

Solution

Balanced equation

2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)

mol AgNO3 :

= 58.5 : 169,87 g/mol

= 0.344

mol CuCl2 :

=21.7 : 134,45 g/mol

= 0.161

mol ratio : coefficient of AgNO3 : CuCl2 :

= 0.344/2 : 0.161/1

= 0.172 : 0.161

CuCl2 as a limiting reagent

mol AgCl :

= 2/1 x 0.161

= 0.322

Mass AgCl :

= 0.322 x 143,32 g/mol

= 46.149 g

mol remains(unreacted) for AgNO3 :

= 0.344-(2/1 x 0.161)

= 0.022

mass AgNO3 remains :

= 0.022 x 169,87 g/mol

= 3.74 g

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Determine the molarity for each of the following solution solutions:

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Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

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Answer:

1.9 × 10² g NaN₃

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Explanation:

Step 1: Write the balanced decomposition equation

2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

Step 2: Calculate the moles of N₂ formed

N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol

We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.

4.2 mol × 28.01 g/mol = 1.2 × 10² g

Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂

The mass ratio of NaN₃ to N₂ is 130.02:84.03.

1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃

Step 4: Calculate the density of N₂

We will use the following expression.

ρ = P × M / R × T

ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L

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