iron is made by a reduction of iron oxide with carbon monoxide. Fe2O3+3CO-2Fe+3CO2. Calculate the mass of iron that can be formed from 126g of iron oxide

The mass of iron formed from 126 grams of iron oxide has been 88.126 grams.

According to the balanced chemical equation, 1 mole of iron oxide reacts with 3 moles of carbon monoxide to form 2 moles of iron and 3 moles of carbon dioxide.

The equation gives:

1 mole iron oxide = 2 moles iron

1 mole iron oxide = 159.69 grams

1 mole of iron = 55.845 grams.

2 moles of iron = 111.69 grams.

159.69 grams Iron oxide = 111.69 grams iron

126 grams iron oxide = 88.126 grams.

The mass of iron formed from 126 grams of iron oxide has been 88.126 grams.

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Mass = 88.12 g

Explanation:

Given data:

Mass of iron oxide = 126 g

Mass of iron formed = ?

Solution:

Chemical equation:

Fe₂O₃ + 3CO    →      2Fe + 3CO₂

Number of moles of iron oxide:

Number of moles = mass/molar mass

Number of moles = 126 g/ 159.69 g/mol

Number of moles = 0.789 mol

Now we will compare the moles of iron with iron oxide.

                      Fe₂O₃           :             Fe

                         1                 :               2

                   0.789              :            2/1×0.789 = 1.578 mol

Mass of iron:

Mass = number of moles ×molar mass

Mass = 1.578 mol × 55.84 g/mol

Mass = 88.12 g

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when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol