Chemistry, 20.12.2020 01:00, missfuture2020p5umxt

In the reaction of aluminum with iron III oxide, if you start with 54.2 grams of aluminum, how many grams of iron III oxide are needed for the reaction to occur?

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answered: mayhem0316oyrjuo

$m_{Fe_2O_3}=160.41gFe_2O_3$

Explanation:

Hello!

In this case, since the reaction between aluminum and iron (III) oxide is:

$2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe$

In such way, since there is 2:1 mole ratio between aluminum (atomic mass = 26.98 g/mol) and iron (III) oxide (molar mass = 159.70 g/mol), we'll be able to compute the mass of the required reactant as shown below:

$m_{Fe_2O_3}=54.2gAl*\frac{1molAl}{26.98gAl}* \frac{1molFe_2O_3}{2molAl}*\frac{159.70gFe_2O_3}{1molFe_2O_3}\\\\m_{Fe_2O_3}=160.41gFe_2O_3$

Best regards!

answered: Guest

when atmosphere interacts with biosphere they focuses on on the sources and sinks of green house gases (ghgs) air pollutants, particular matters, water and energy between the earth's surface and the atmosphere. the biosphere affects the climate. the atmosphere and biosphere are in constant interaction with each other, and all living things rely on gases in the atmosphere for life. organisms such as animals take in oxygen gas form the atmosphere to use in every cell of their body for smooth functioning of daily routine exercises.

What happens when the atmosphere interacts with biosphere

answered: Guest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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In the reaction of aluminum with iron III oxide, if you start with 54.2 grams of aluminum, how many...