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3 years ago

_Fe2O3 + 2CO —> _Fe + _CO2

Chemistry

1 answer:

soldi70 [24.7K]3 years ago

8 0

<h3>Answer:</h3>

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
Therefore; the balanced chemical equation for the reaction is;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

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A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc

Paul [167]

Answer : The volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of $O_2$ gas = (740-22.4) torr = 717.6 torr

$P_2$ = final pressure of $O_2$ gas at STP= 760 torr

$V_1$ = initial volume of $O_2$ gas = 280 mL

$V_2$ = final volume of $O_2$ gas at STP = ?

$T_1$ = initial temperature of $O_2$ gas = $25^oC=273+25=298K$

$T_2$ = final temperature of $O_2$ gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get:

$\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}$

$V_2=242.2mL=0.2422L$

Therefore, the volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

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3 years ago

Write a chemical equation for the following decomposition reaction.

Aneli [31]

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</span><span>we know that the heat supplied to decompose the compound.In the result the product H2O is assumed to be in the vapor state so that is gas.

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3 years ago

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3 years ago

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3 years ago

Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 %

Mademuasel [1]

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

$Moles\ Ca = \frac{38.7g}{40.08g/mol} =0.966\ mol\\\\Moles\ P = \frac{19.9g}{30.97g/mol} =0.643\ mol\\\\Moles\ O = \frac{41.2g}{16.00g/mol} =2.58\ mol$

Step 2: Calculate the molar ratio

$C = \frac{0.966}{0.643} =1.50\\\\P = \frac{0.643}{0.643} = 1.00\\\\O = \frac{2.58}{0.643} =4.00$

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

7 0

3 years ago