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viva [34]
4 years ago
10

How many moles of aluminum are needed to react completely with 1.2 mol of feo

Chemistry
1 answer:
Amanda [17]4 years ago
3 0

Answer:

0.8 mol.

Explanation:

  • The balanced equation for the reaction between Al and FeO is represented as:

<em>2Al + 3FeO → 3Fe + Al₂O₃,</em>

It is clear that 2 mol of Al react with 3 mol of FeO to produce 3 mol of Fe and 1 mol of Al₂O₃.

<em><u>Using cross multiplication:</u></em>

2 mol of Al needs → 3 mol of FeO, from stichiometry.

??? mol of Al needs → 1.2 mol of FeO.

∴<em> The no. of moles of Al are needed to react completely with 1.2 mol of FeO </em>= (2 mol)(1.2 mol)/(3 mol) = <em>0.8 mol.</em>

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An organic compound contains , , , and . Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g and 0.0861 g .
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The question displayed below shows the missing information which therefore completes the question.

An organic compound contains C, H, N and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g of CO2 and 0.0861 g of H2O. A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method. The compound is first reacted by passage over hot: The product gas is then passed through a concentrated solution of to remove the. After passage through the solution, the gas contains and is saturated with water vapor. At STP, 38.9 mL of dry N2 was obtained. In a third experiment, the density of the compound as a gas was found to be 2.86 g/L at 127°C and 256 torr. What are the empirical and molecular formulas of the compound? (Enter the elements in the order: C, H, N, O.)

Answer:

the empirical formula = \mathbf {C_3H_6O_{12}N}

the molecular formula = \mathbf {C_3H_6O_{12}N}

Explanation:

From the given information:

\bigg ( 0.2587 \ g \ of CO_2 \bigg) \times \dfrac{1 \ mol \ of CO_2}{44 \ of \ CO_2} \times \dfrac{1 \ mol \ of \  C}{1 \ mol \ of CO_2}

= 0.00588 \ mol \ of \ C \times \dfrac{12.01 \ g \ of \ C}{1 \ mol \ of \ C }

= 0.0706g of C

\bigg ( 0.0861\ g \ of H_2O \bigg) \times \dfrac{1 \ mol \ of H_2O}{18.02 \ g  \ of \ H_2O} \times \dfrac{2 \ mol \ of \  H}{1 \ mol \ of H_2O}

=0.0096 \ mol \times \dfrac{1.008 \ g \ of \ H}{1 mol \ H}

0.0097g of H

Given that N2 at STP = 1 atm, 273 K and V = 0.0389 L

PV = nRT

n = PV/RT

n = \dfrac{1 \ atm \times 0.0389 \ of \ H_2}{0.0821 \ L.atm /mol.K \times 273 \ K }

n = 0.00173 mol of N2

The oxygen in the sample = The total grams in sample -  gram in H - gram in C

The oxygen in the sample = 0.1023 g - 0.0097 g - 0.706 g

The oxygen in the sample = 0.022 g of O

The number of  moles of O_2 = \dfrac{0.02}{16}

= 0.001375 mol of O

O \ in \ product = (0.00588 \ mol \ of \ C ) \times \dfrac{2 \ mol \ of \ O }{1 \ mol \ of \ C }+ \bigg ( 0.0096 \ mol \ of \ H ) \times \dfrac{1 \ mol \ of  \ O }{1 \ mol \ of \ H}

O in product = 0.02136 mol of O

∴

we are meant to divide the moles of each compound by the smallest number of  moles; we have:

C = \dfrac{0.00588}{0.00173} \simeq 3

H = 0.0096 = \dfrac{0.0096}{0.00173} \simeq 6

O = 0.0199= \dfrac{0.0199}{0.00173} \simeq 12

N = 0.00173= \dfrac{0.00173}{0.00173} \simeq 1

Thus; the empirical formula = \mathbf {C_3H_6O_{12}N}

To estimate the molecular formula;  we have:

MM = \dfrac{dRT}{P}

MM = \dfrac{2.80 \ g/ L \times 0.0821 \ L.atm /mol.K \times 400 \ K }{0.337 \ atm}

MM = 272.86 g/mol

Also; the molar mass of \mathbf {C_3H_6O_{12}N} = 248 g/mol

∴

= \dfrac{272.86 \ g/mol}{248 \ g/mol}

=1

Thus; we can conclude that empirical formula as well as the molecular formula are the same.

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