Thus problem is providing us with the mass of iron (III) oxide as 12.4 g so the moles are required and found to be 0.0776 mol after the calculations:

<h3>Mole-mass relationships:</h3>

In chemistry, we use mole-mass relationships in order to calculate grams from moles and vice versa. In this case, since we are given the mass of iron (III) oxide as 12.4 g one can calculate the moles by firstly quantifying its molar mass:

$Fe_2O_3\rightarrow 2*55.85 g/mol+3*16.00 g/mol=159.7g/mol$

Then, we prepare a conversion factor in order to cancel out the grams and thus, get moles:

$12.4gFe_2O_3*\frac{1molFe_2O_3}{159.7gFe_2O_3} \\\\=0.0776molFe_2O_3$

Learn more about mole-mass relationships: brainly.com/question/18311376

8 0

2 years ago

How do you convert between the volume of a gas at STP and the number of moles of the gas?

Naya [18.7K]

Standard temperature and pressure (STP) means a temperature of 0°c and a pressure of 1 atmosphere (atm). The molar gas volume is used to convert between the number of moles of a gas and the volume of the gas at STP. One mole of a gas occupies a volume of 22400 cm³ or 22.4 liters at STP according to the molar gas volume.

6 0

3 years ago

DUE TOMORROW!!! 15 POINTS

Lunna [17]

Answer:

A. write balanced chemical equation (including states), for this process.

Explanation:

Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.

Hydrocarbon combustions always involve

[some hydrocarbon] + oxygen --> carbon dioxide + steam.

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side:

C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

Balance hydrogen, six on each side:

C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

Now, we have fifteen oxygens on the right and O2 on the left.

Two ways to deal with that. We can use a fraction:

C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

Or, if you prefer to have whole number coefficients, double everything

to get rid of the fraction:

2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

With the SATP states thrown in...

C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)

6 0

3 years ago