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1 year ago

A sample of the mineral hematite Iron (III) oxide has a mass of 12.4g. How many moles of the mineral are present?

1 answer:

8 0

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

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In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what

melomori [17]

Answer:

$Y=65.7\%$

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

$n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2$

Thus, we compute the percent yield as:

$Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%$

Best regards.

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