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1 year ago

How many moles of silver can be produced from silver nitrate from 1 mole of zinc?

Chemistry

1 answer:

jasenka [17]1 year ago

4 0

Answer:

Answer: 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

Explanation:

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

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

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = 1/2 x 6.5 = 3.25 moles of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = 2/2 x 6.5 = 6.5moles of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

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2 years ago

2 A certain gas of 25 g at 25°c and 0.65 atm occupies a volume of 23.52L Determine the molecule mass of the gas.

denpristay [2]

Answer:

${ \bf{PV= \frac{m}{M} RT}} \\ \\ { \tt{(0.65 \times 23.52) = \frac{25}{M} \times 0.081 \times (25 + 273)}} \\ \\ M = 39.5 \: g$

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3 years ago

Calculate the area of a 3.0 inch by 5.0 inch index card in square millimeters (mm). (You can look up the formula for the area of

meriva

Answer:

The area of the given rectangular index card = 9677.4 mm²

Explanation:

Area is defined as the space occupied by a two dimensional shape or object. The SI unit of area is square metre (m²).

The area of a rectangle (A) = length (l) × width (w)

Given dimensions of the rectangle: Length (l) = 5.0 inch, Width (w) = 3.0 inch

Since, 1 inch = 25.4 millimetres (mm)

Therefore, l = 5 × 25.4 = 127 mm, and w = 3 × 25.4 = 76.2 mm

Therefore, the area of the given rectangular index card = A= l × w = 127 mm × 76.2 mm = 9677.4 mm²

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2 years ago

Cho 4g CuO vào dung dịch axit clohidric 10% thì phản ứng vừa đủ.

Sholpan [36]

Answer:

Explanation:

a. CuO+ 2HCl⇒CuCl2+ H2O

b. $n_{CuO}$ = $\frac{4}{80}$ = 0,05 (mol)

⇒ $n_{CuCl2}$ = $n_{CuO}$ =0,05 mol

⇒ $m_{CuCl2}$ = 0,05×135=6,75 (g)

c. $n_{HCl}$ =2× $n_{CuO}$ =0,1 (mol)

⇒ $m_{HCl}$ = 0,1×36,5= 3,65 (g)

⇒ $m_{dd HCl}$ = $\frac{m_{HCl}}{10}$ ×100=36,5 (g)

⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2= $\frac{m_{CuCl2} }{m_{dd HCl+ m_{CuO} } }$ ×100= $\frac{6,75}{36,5+4}$ ×100≈ 16,67%

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3 years ago

Determine the formula weights of each of the following compounds.Part A) Nitrous oxide, N2O, known as laughing gas and used as a

densk [106]

Answer:

See explanation

Explanation:

a) Nitrous oxide (N2O) has a molar mass of 44.014 amu. It has 2 nitrogen atoms each with a mass of 14.007 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 44.014 amu ) *100% = 63.6%

Percentage oxygen = (16 amu/44.014 amu) * 100% = 36.4 %

63.6% + 36.4% = 100%

b) Benzoic acid (C7H6O2) has a molar mass of 122.13 amu. It has 6 hydrogen atoms each with a mass of 1.01 amu; it has 7 carbon atoms each with a mass of 12.01amu and 2 oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (6*1.01 amu / 122.13 amu)*100% = 4.96%

Percentage carbon = (7*12.01 amu/ 122.13 amu)*100% = 68.8%

Percentage oxygen = (2*16 amu/ 122.13 amu) *100% = 26.2%

c) Magnesium hydroxide (Mg(OH)2) has a molecular mass of 58.32 amu. It has 2 hydrogen atoms each with a mass of 1.01 amu; it has 1 magnesium atom with a mass of 24.3 amu and two oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (2*1.01 amu/ 58.32 amu) *100% = 3.46 %

Percentage magnesium = (24.3 amu/58.32 amu)*100% = 41.7%

Percentage oxygen = (2*16 amu/58.32 amu)*100% = 54.9%

d) Urea CO(NH2)2 has a molecular mass of 60.064 amu. It has 2 Nitrogen atoms each with a mass of 14.007 amu, 4 hydrogen atoms each with a mass of 1.01 amu,1 carbon atom with a mass of 12.01 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 60.064amu)*100% = 46.6%

Percentage hydrogen = (4*1.01 amu/60.064amu)*100% = 6.72%

Percentage carbon = (12.01 amu/60.064amu)*100% = 20.0%

Percentage oxygen = (16 amu/60.064amu)*100% = 26.6%

e) Osopentyl acetate (C7H14O2) has a molecular mass of 130.2 amu. It has 14 hydrogen atoms each with a mass of 1.01 amu,7 carbon atoms each with a mass of 12.01 amu and 2 oxygen atom with a mass of 16.0 amu.

Percentage hydrogen = (14*1.01 amu/130.2 amu)*100% = 10.8%

Percentage carbon = (7*12.01 amu/130.2 amu)*100% = 64.6%

Percentage oxygen =(2*16 amu/130.2 amu)*100% = 24.6%

7 0

3 years ago