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If 2.45 g of iron are placed in 1,5 L of 0.25M HCl, how many grams of FeCl2 are obtained? Identify the limiting and excess react

lys-0071 [83]

Answer:

$m_{FeCl_2}=0.652gFeCl_2$

Explanation:

Hello there!

In this case, according to the given information and the chemical reaction, whereby iron and hydrochloric acid react in a 1:2 mole ratio, it is firstly necessary to calculate the moles of iron (II) chloride from each reactant in order to figure out the limiting reactant:

$n_{FeCl_2}=2.45gFe*\frac{1molFe}{55.845gFe}*\frac{1molFeCl_2}{1molFe}=0.0439molFeCl_2\\\\ n_{FeCl_2}=1.5L*0.25\frac{molHCl}{L} *\frac{1molHCl}{36.46gHCl}*\frac{1molFeCl_2}{2molHCl}=0.00514molFeCl_2$

In such a way, we infer the maximum moles of FeCl2 product are yielded by HCl, for which it is the limiting reactant. Finally, we calculate the grams of product by using its molar mass as shown below:

$m_{FeCl_2}=0.00514molFeCl_2*\frac{126.75gFeCl_2}{1molFeCl_2} \\\\m_{FeCl_2}=0.652gFeCl_2$

Regards!

7 0

3 years ago

In the reaction between copper and silver nitrate, 95.3 g of silver and 82.9g of copper nitrate are produced. How much of each r

Nastasia [14]

Reaction

2AgNO₃ + Cu ⇒ Cu(NO₃)₂ + 2Ag

mol silver (Ag): 95.3 : 108 g/mol = 0.882

mol copper nitrate (Cu(NO₃)₂): 82.9 : 187.5 g/mol = 0.441

mol Ag = 2 x mol Cu(NO₃)₂, so 0.441 being the mole basis of the reactants

mol Cu = 0.441

mass Cu = 0.441 x 63.5 g/mol = 28 g

mol AgNO₃ = 0.882

mass AgNO₃ = 0.882 x 170 g/mol = 149.94 g

5 0

3 years ago

A metamorphic rock can also be thought of as a rock that changes. What causes the rock to change?

Nana76 [90]

Answer:

its ethier b or c

Explanation:

5 0

3 years ago

Read 2 more answers

A student working in the laboratory produces 6.81 grams of calcium oxide, CaO, from 20.7 grams of calcium

xz_007 [3.2K]

Answer:

A. Theoretical yield of CaO is 11.59 g

B. Percentage yield of CaO = 58.76%

Explanation:

The following data were obtained from the question:

Mass of CaCO₃ = 20.7 g

Actual yield of CaO = 6.81 g

Theoretical yield of CaO =?

Percentage yield of CaO =?

The equation for the reaction is given below:

CaCO₃ —> CaO + CO₂

Next, we shall determine the mass of CaCO₃ that decomposed and the mass of CaO produced from the balanced equation. This can be obtained as follow:

Molar mass of CaCO₃ = 40 + 12 + (3×16)

= 40 + 12 + 48

= 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

Molar mass of CaO = 40 + 16 = 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

SUMMARY:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

A. Determination of the theoretical yield of CaO.

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Therefore, 20.7 g of CaCO₃ will decompose to produce =

(20.7 × 56)/100 = 11.59 g of CaO.

Thus, the theoretical yield of CaO is 11.59 g

B. Determination of the percentage yield.

Actual yield of CaO = 6.81 g

Theoretical yield of CaO = 11.59 g

Percentage yield of CaO =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 6.81/11.59 × 100

Percentage yield of CaO = 58.76%

4 0

3 years ago

3) How Many moles are in 0.500 milligrams of the “emergency” hormone, adrenaline, whose formula

Yuki888 [10]

To determine the total number of moles that are in the hormone, you would simply convert that to grams and then use the molar mass if the formula to solve for the total number of moles in the compound.

5 0

4 years ago