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3 years ago

How many grams of Fe3O4 are required to react completely with 300 grams of H2?

Chemistry

1 answer:

vekshin13 years ago

8 0

Answer:

2023.04 g

Explanation:

Magnetite reacts with hydrogen to produce Iron metal and steam. Steam instead of water is produced as the reaction occurs at temperatures above the boiling point of water.

Fe₃O₄ + 4 H₂ → 3 Fe +4 H₂O

From the equation, 1 mole of Fe₃O₄ reacts with 4 moles of H₂.

69.76 grams of H₂ has the following number of moles.

Number of moles= mass/RAM

=69.76/2

=34.88 moles.

The reaction ratio of Fe₃O₄ to H₂ is 1:4

Thus number of moles of magnetite= (1×34.88)/4

=8.72 moles.

Mass= moles × molecular weight

=8.72 moles × (56×3+16×4)

=2023.04 grams

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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.

Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

$Mg+2HCl\rightarrow MgCl_2+H_2$

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

$n_{H_2}^{by\ HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl} =0.0284molH_2\\\\n_{H_2}^{by\ Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2$

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

$m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2$

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

$m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg} =0.690gMg$

Thus, the mass of excess magnesium turns out:

$m_{Mg}^{excess}=2.38g-0.690g=1.69gMg$

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

$Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%$

Best regards!

8 0

3 years ago

Identify the correct coefficients to balance the redox reaction with the lowest possible integer coefficients.

Monica [59]

Answer:

$\rm 3\; Ag^{1+} + 1\; Al \to 1\; Al^{3+} + 3\; Ag$ .

Explanation:

Electrons are conserved in a chemical equation.

The superscript of $\rm Ag^{1+}$ indicates that each of these ions carries a charge of $+1$ . That corresponds to the shortage of one electron for each $\rm Ag^{+}$ ion.

Similarly, the superscript $+3$ on each $\rm Al^{3+}$ ion indicates a shortage of three electrons per such ion.

Assume that the coefficient of $\rm Ag^{+}$ (among the reactants) is $x$ , and that the coefficient of $\rm Al^{3+}$ (among the reactants) is $y$ .

$\rm \mathnormal{x}\; Ag^{1+} + ?\; Al \to \mathnormal{y}\; Al^{3+} + ?\; Ag$ .

There would thus be $x$ silver ( $\rm Ag$ ) atoms and $y$ aluminum ( $\rm Al$ ) atoms on either side of the equation. Hence, the coefficient for $\rm Al\!$ and $\rm Ag\!$ would be $y\!$ and $x\!$ , respectively.

$\rm \mathnormal{x}\; Ag^{1+} + \mathnormal{y}\; Al \to \mathnormal{y}\; Al^{3+} + \mathnormal{x}\; Ag$ .

The $x$ $\rm Ag^{1+}$ ions on the left-hand side of the equation would correspond to the shortage of $x$ electrons. On the other hand, the $y$ $Al^{3+}$ ions on the right-hand side of this equation would correspond to the shortage of $3\, y$ electrons.

Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of $x$ electrons, the right-hand side should also be $x\!$ electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of $3\, y$ electrons. These two expressions should have the same value. Therefore, $x = 3\, y$ .