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aksik [14]
3 years ago
8

Find the grams of iron in 79.2 g of Fe2O3.

Chemistry
1 answer:
Setler79 [48]3 years ago
3 0
First you find the RFM (Relative Formula Mass) of Fe2O3
Fe2 = 56 x 2
= 112
O3 = 16 x 3
= 48

112 + 48 = 160

Moles = mass/RFM
           = 79.2/160
           =0.495

Then you rearrange the formula
Mass= moles x RFM
        = 0.495 x 112
        = 55.44 g
(To find mass of Iron have to use RFM of total iron in the substance example Fe2 ^^)


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<u>Answer:</u> The mass of H_2O produced is 2.52 g

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For hexane:</u>

Given mass of hexane = 1.72 g

Molar mass of hexane = 86.18 g/mol

Putting values in equation 1, we get:

\text{Moles of hexane}=\frac{1.72g}{86.18g/mol}=0.020mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 8.0 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{8.0g}{32g/mol}=0.25mol

The chemical equation for the combustion of hexane follows:

2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O

By stoichiometry of the reaction:

If 2 moles of hexane reacts with 19 moles of oxygen gas

So, 0.020 moles of hexane will react with = \frac{19}{2}\times 0.020=0.19mol of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hexane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of hexane produces 14 moles of H_2O

So, 0.020 moles of hexane will produce = \frac{14}{2}\times 0.020=0.14mol of H_2O

We know, molar mass of H_2O = 18 g/mol

Putting values in above equation, we get:

\text{Mass of }H_2O=(0.14mol\times 18g/mol)=2.52g

Hence, the mass of H_2O produced is 2.52 g

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The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh
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When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

150.0 mL \times \frac{1.02g}{mL}  = 153 g

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J  }{\frac{4.184J}{g.\° C }  \times 153g} = 5.87 \° C

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

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