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2 years ago

Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:

Chemistry

1 answer:

Lerok [7]2 years ago

7 0

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

<em>%yield = 88.5%</em>

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

Read 2 more answers

A mixture was found to contain 1.05 g if silicone dioxide, 0.69 g cellulose, 1.82 g of calium carbonate. what percent of calcium

djverab [1.8K]

1,05+ 0,69 + 1,82 = 3,56g of mixture

%CaCO₃: 1,82/3,56×100% = 0,5112×100% = 51,12%

5 0

2 years ago

A molecule made of hydrogen and carbon with Mr of 56 has the following composition: carbon 85.7%; hydrogen 14.3%. Calculate the

Veseljchak [2.6K]

Answer:

Assume that 100 grams of C2H4 is present. This means that there are 85.7 grams of carbon and 14.3 grams of hydrogen.

Convert these weights to moles of each element:

85.7 grams carbon/12 grams per mole = 7 moles of carbon.

14.3 grams hydrogen/1 gram per mole = 14 moles of hydrogen.

Divide by the lowest number of moles to obtain one mole of carbon and two moles of hydrogen.

Since we know that there cannot be a stable CH2 molecule, multiply by two and you have C2H4 which is ethylene - a known molecule.

The secret is to convert the percentages to moles and find the ration of the constituents.

5 0

2 years ago

How do the properties of the elements change as you move across a period on a periodic table from left to right?

Tomtit [17]

B) The elements become less reactive.

8 0

3 years ago

Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu

MAVERICK [17]

Answer : The initial temperature of system 2 is, $19.415^oC$

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

$-q_1=q_2$

$m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)$

The mass remains same.

where,

$C_1$ = heat capacity of system 1 = 19.9 J/mole.K

$C_2$ = heat capacity of system 2 = 28.2 J/mole.K

$T_f$ = final temperature of system = $30^oC=273+30=303K$

$T_1$ = initial temperature of system 1 = $45^oC=273+45=318K$

$T_2$ = initial temperature of system 2 = ?

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

$-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K$

$T_2=292.415K$

$T_2=292.415-273=19.415^oC$

Therefore, the initial temperature of system 2 is, $19.415^oC$

8 0

3 years ago