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Reil [10]
4 years ago
9

Cu+2AgNO

Chemistry
1 answer:
baherus [9]4 years ago
6 0

Answer:

Mass of Ag produced = 64.6 g

Note: the question is, how many grams of Ag is produced from 19.0 g of Cu and 125 g of AgNO3

Explanation:

Equation of the reaction:

Cu + 2AgNO3 ---> 2Ag + Cu(NO3)2

From the equation above, 1 mole of Cu reacts with 2 moles of AgNO3 to produce 2 moles of Ag and 1 mole of Cu(NO3)2.

Molar mass of the reactants and products are; Cu = 63.5 g/mol, Ag = 108 g/mol, AgNO3 = 170 g/mol, Cu(NO3)2 = 187.5 g/mol

To determine, the limiting reactant;

63.5 g of Cu reacts with 170 * 2 g of AgNO3,

19 g of Cu will react with (340 * 19)/63.5 g of AgNO3 =101.7 g of AgNO3.

Since there are 125 g of AgNO3 available for reaction, it is in excess and Cu is the limiting reactant.

63.5 g of Cu reacts to produce 108 * 2 g of Ag,

19 g of Cu will react to produce (216 * 19)/63.5 g of Ag = 64.6 g of Ag.

Therefore mass of Ag produced = 64.6g

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How many milliliters of 2.19 M H2SO4 are required to react with 4.75 g of solid containing 21.6 wt% Ba(NO3)2 if the reaction is
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Answer:

1.7927 mL

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Mass of Ba(NO_3)_2 = \frac {21.6}{100}\times 4.75\ g = 1.026 g

Molar mass of Ba(NO_3)_2 = 261.337 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.026\ g}{261.337\ g/mol}

Moles= 0.003926\ mol

Considering the reaction as:

Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3

1 moles of Ba(NO_3)_2 react with 1 mole of H_2SO_4

Thus,

0.003926 mole of Ba(NO_3)_2 react with 0.003926 mole of H_2SO_4

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Also, considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity = 2.19 M

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2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}

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