For the following reaction: Fe2O3 + 3CO --> 2Fe + 3CO2 how many grams of Fe2O3 is needed to produce 111 g of Fe?

1 answer:

4 0

You need 158.70 grams of Fe2O3 to produce 111 grams of Fe. This is calculated by using the molar masses and stoichiometric relationship of the two compounds.

Solution:

MM Fe = 55.845 g/mol
MM Fe2O3 = 159.69 g/mol
Fe: Fe2O3 = 2 mol:1 mol

11 g FE (1 mol Fe/55.845 g Fe) (1 mol Fe2O3/2 mol Fe) (159.69 g Fe2O3 / 1 mole Fe2O3) = 158.70 grams Fe2O3

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A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$