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weeeeeb [17]
3 years ago
14

A student wants to reclaim the iron from an 18.0 gram sample of iron (III) oxide, which occurs according to the reaction below:

how many grams of iron can be reclaimed.
Chemistry
1 answer:
geniusboy [140]3 years ago
4 0

Answer:

12.6 g

Explanation:

Step 1: Write the balanced decomposition reaction

2 Fe₂O₃ ⇒ 4 Fe + 3 O₂

Step 2: Calculate the moles corresponding to 18.0 g of Fe₂O₃

The molar mass of Fe₂O₃ is 159.69 g/mol.

18.0 g × 1 mol/159.69 g = 0.113 mol

Step 3: Calculate the moles of Fe formed from 0.113 moles of Fe₂O₃

The molar ratio of Fe₂O₃ to Fe is 2:4. The moles of Fe formed are 4/2 × 0.113 mol = 0.226 mol

Step 4: Calculate the mass corresponding to 0.226 moles of Fe

The molar mass of Fe is 55.85 g/mol.

0.226 mol × 55.85 g/mol = 12.6 g

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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.
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Answer:

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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!

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