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4 years ago

6

Change 0.00765 kL into mL

2 answers:

bagirrra123 [75]4 years ago

4 0

Answer:

7650

Explanation:

formula- multiply the volume value by 1e+6

kvasek [131]4 years ago

3 0

Answer

7650

multiply the volume value by 1+6

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4 years ago

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What mass of iron(II) oxide must be used in the reaction given by the equation below to release 44.7 kJ? 6FeO(s) + O2(g) => 2

zavuch27 [327]

<u>Answer:</u> The mass of iron (II) oxide that must be used in the reaction is 30.37

<u>Explanation:</u>

The given chemical reaction follows:

$6FeO(s)+O_2(g)\rightarrow 2Fe_3O_4(s);\Delta H^o=-635kJ$

By Stoichiometry of the reaction:

When 635 kJ of energy is released, 6 moles of iron (II) oxide is reacted.

So, when 44.7 kJ of energy is released, $\frac{6}{635}\times 44.7=0.423mol$ of iron (II) oxide is reacted.

Now, calculating the mass of iron (II) oxide by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of iron (II) oxide = 0.423 moles

Molar mass of iron (II) oxide = 71.8 g/mol

Putting values in above equation, we get:

$0.423mol=\frac{\text{Mass of FeO}}{71.8g/mol}\\\\\text{Mass of FeO}=(0.423mol\times 71.8g/mol)=30.37g$

Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37

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