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3 years ago

iron will react with oxygen to produce Fe2O3. how many grams of Fe2O3 will he produced if 0.18 mil of Fe reacts?

Chemistry

1 answer:

nasty-shy [4]3 years ago

6 0

Answer:

14.4g

Explanation:

First, we need to write a balanced equation for the reaction between Fe and O2 to produce Fe2O3. This is illustrated below:

4Fe + 3O2 —> 2Fe2O3

From the balanced equation,

4moles of Fe produced 2moles of Fe2O3.

Therefore, 0.18mol of Fe will produce = (0.18x2) /4 = 0.09mol of Fe2O3.

Now we need to find the mass present in 0.09mol of Fe2O3. This can be achieved by doing the following:

Molar Mass of Fe2O3 = (56x2) + (16x3) = 112 + 48 = 160g/mol

Number of mole of Fe2O3 = 0.09mol

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Fe2O3 = 0.09 x 160 = 14.4g

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Explanation:

There is a mistake in the statement. Correct form is described below:

<em>Using the above data table and graph, calculate the total energy in Joules required to raise the temperature of 15 grams of ice at -5.00 °C to water at 35 °C. </em>

The total energy needed to raise the temperature is the combination of latent and sensible heats, all measured in joules, and represented by the following model:

$Q = m\cdot [c_{i} \cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})]$ (1)

Where:

$m$ - Mass of the sample, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of the sample, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of water, in degrees Celsius.

$Q$ - Total energy, in joules.

If we know that $m = 15\,g$ , $c_{i} = 2.06\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.184\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334.72\,\frac{J}{g}$ , $T_{1} = -5\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ and $T_{3} = 35\,^{\circ}C$ , then the final energy to raise the temperature of the sample is:

$Q = (15\,g)\cdot \left[\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (5\,^{\circ}C)+ 334.72\,\frac{J}{g} + \left(4.184\,\frac{J}{g\cdot ^{\circ}C}\right)\cdot (35\,^{\circ}C) \right]$

$Q = 7371.9\,J$

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Answer:

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