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NemiM [27]
2 years ago
8

If you start with 4 moles of iron and 3 moles of oxygen to produce iron oxide, what is the limiting reagent? (You will need to b

alance the equation first.) Fe + O2 -> Fe2O3
A. they are equal
B.Fe
C. O2
D.Fe2O3​
Chemistry
1 answer:
Marianna [84]2 years ago
8 0

The correct statement regarding the limiting reactant for the reaction between 4 moles of iron and 3 moles of oxygen is: They are both equal. (Option A)

<h3>Balanced equation</h3>

We'll begin by writing the balanced equation for the reaction between iron and oxygen. This is given below:

4Fe + 3O₂ --> 2Fe₂O₃

SUMMARY

From the balanced equation above,

4 moles of Fe reacted with 3 moles of O₂

<h3>How to determine the limiting reactant</h3>

The limiting reactant is the reactant that is consumed completly in a chemical reaction.

The limiting reactant can be obtain as illustrated below:

From the balanced equation above,

4 moles of Fe reacted with 3 moles of O₂

From the illustration above, we can see that 3 moles of Fe required 4 moles of O₂ for complete reaction.

Thus, Fe and O₂ are both the limiting reactant.

Therefore, we can conclude that Fe and O₂ are equal

Learn more about stoichiometry:

brainly.com/question/11587316

#SPJ1

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If 37.5 mL of 0.100 M calcium chloride reacts completely with aqueous silver nitrate, what is the mass of AgCl (MM
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Answer:

OPTION C is correct

C) 1.07 g

Explanation:

CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)3(aq)

But we know molarity molarity= number of moles of solute/ volume of the solution

M= n/V

From the equation above

number of moles of Cacl2 = (37.5 ×0 .100 × 10^-3) = 0.00375 moles.

Then

1 mole of Cacl2 yields 2 moles of Agcl2

0.00375 moles of Cacl2 will produce let say Y.

Y= (0.00375 ×2)/1

= 0.0075 moles.

Number of moles of Agcl2 = mass /molar mass of Agcl

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Then mass = 178 ×0.0075 = 1.047

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Molecular mass of C₂H₄ is,

M = 2×12 + 4×1 g/mol

M = 28 g/mol

Moles of C₂H₄ in 5.6 g of C₂H₄ :

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n = 0.2 mol

Now, 1 mol of C₂H₄ contains 2 moles of carbon.

So, number of moles of carbon are :

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So, number of carbon atoms are :

N = 0.4 \times 6.022\times 10^{23} \\\\N = 2.409 \times 10^{23}

Hence, this is the required solution.

4 0
3 years ago
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