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ahrayia [7]
3 years ago
15

A chemical company makes pure silver by reacting silver nitrate with Zinc. the company needs to make 800 grams of pure silver fo

r a client. they have 500 grams of zinc and 1500 grams of silver nitrate will the make enough silver to fill the order
Chemistry
1 answer:
kirill115 [55]3 years ago
3 0

Answer:

Yes. Since they have more than the required reactants.

Explanation:

Zinc reduces silver nitrate to silver metal according to the following equation.

2AgNO₃ + Zn → 2Ag + Zn(NO₃)₂

From the equation 2 moles of AgNO₃ produce 2 moles of silver.

If we consider the zinc to be used number of moles=mass/RAM

RAM of zinc =65.38

No. of moles=500/65.38

=7.6476moles

To produce 800 grams of silver they require:

800/107.868 moles=7.4165 moles of silver nitrate since the ratio of silver nitrate to silver produced is 1:1

The number of moles of silver nitrate available=1500/169.872

=8.83 moles.

As the amount of reactants available is more than the required the company will make it in producing the 800 grams of silver required.

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Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.
Bezzdna [24]

Answer:

\large \boxed{\text{ 0.066 mol/L}}

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ:       58.44  

            NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g:     0.245

V/mL:                 50.

c/mmol·mL⁻¹:       0.0180

2. Calculate the moles of each reactant  

\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}

3. Identify the limiting reactant  

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:  

The molar ratio of NaCl to AgCl is 1:1.

\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}

From AgNO₃:  

The molar ratio of AgNO₃ to AgCl is 1:1.  

\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

                   Ag⁺(aq)  +  Cl⁻(aq) ⟶ AgCl(s)

 I/mmol:      0.900        4.192            0

C/mmol:    -0.900       -0.900        +0.900

E/mmol:      0                3.292          0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}

8 0
3 years ago
A flashlight has a resistance of 2.4 . What voltage is applied by the batteries if the current in the circuit is
hammer [34]

Answer : The voltage applied by the batteries is, 6.0 V

Solution : Given,

Resistance of flashlight = 2.4 ohm

Current in the circuit = 2.5 Ampere

Formula used :

V=IR

where,

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I = current in the circuit

R = resistance of light

Now put all the given values in the above formula, we get

V=(2.5A)\times (2.4ohm)=6volt=6V

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3 years ago
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