The elements in the periodic table are arranged in order of

If you start with 227.8 grams of iron and 128 grams of oxygen to produce iron oxide, what is the limiting reagent?

agasfer [191]

The balanced chemical equation between iron and oxygen to produce iron (III) oxide is,

$4Fe(s) + 3O_{2}(g) ---> 2Fe_{2}O_{3}(s)$

Mass of Fe = 227.8 g

Moles of Fe = $227.8 g Fe * \frac{1 mol Fe}{55.85 g Fe} = 4.079 mol Fe$

Mass of oxygen = 128 g

Moles of $O_{2} = 128 g O_{2}*\frac{1 mol O_{2}}{32 g O_{2}}= 4mol O_{2}$

Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.

Mass of iron (III) oxide produced from Iron = $4.079 mol Fe * \frac{2 mol Fe_{2}O_{3}}{4 mol Fe} *\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 325.7 g Fe_{2}O_{3}$

Mass of iron (III) oxide produced from oxygen= $4 mol O_{2}*\frac{2 molFe_{2}O_{3}}{3 mol O_{2}}*\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 425.84 g Fe_{2}O_{3}$

Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.

7 0

2 years ago

Pleas help with 2 and 4 for brainliest

Snezhnost [94]

mass of pentane : = 30.303 g

moles of Al₂(CO₃)₃ : = 0.147

<h3>Further explanation</h3>

Given

1. Reaction

C₅H₁₂+8O₂→6H₂O+5CO₂.

45.3 g water

2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂

37.2 MgCO₃

Required

mass of pentane

moles of Al₂(CO₃)₃

Solution

1. mol water = 45.3 : 18 g/mol = 2.52

From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :

= 1/6 x mol H₂O

= 1/6 x 2.52

= 0.42

Mass pentane :

= mol x MW

= 0.42 x 72.15 g/mol

= 30.303 g

2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44

mol Al₂(CO₃)₃ :

= 1/3 x mol MgCO₃

= 1/3 x 0.44

= 0.147

4 0

1 year ago

In the number 48.93, which digit is estimated?

Fudgin [204]

The answer is 3.

Explanation:
It’s the last number and it can’t be 9 because then it would be 48.9 and no 3.

7 0

2 years ago

A train in Japan can travel 813.5 miles in 5 hours

Anastaziya [24]

Answer:

162.7miles/hr

Explanation:

Given parameters:

Distance covered by the train = 813.5miles

Time taken = 5hours

Unknown:

Speed of the train = ?

Solution:

Speed is a physical quantity.

It is mathematically expressed as;

Speed = $\frac{distance}{time}$

So, input parameters and solve;

Speed = $\frac{813.5}{5}$ = 162.7miles/hr

8 0

2 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

2 years ago