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3 years ago

How many moles of iron can be produced from 4.20 moles of iron (III) oxide?

Chemistry

1 answer:

Kay [80]3 years ago

6 0

Hope this helps, have a nice day ahead!

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Two molecules of one reactant combine with 3 molecules of another to produce 5 molecules of a product.

maksim [4K]

Answer:

A = 2A + 3B → 5C

Explanation:

The two molecule of A and three molecules of B will react to form the five molecules of C.

2A + 3B → 5C

Other options are incorrect because,

B = A₂ + B₃ → C₅

in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.

C = 2A + 5B → 3C

in this reaction two molecules of A and five molecules of B combine to form three molecule of C.

D = A₂ + B₃ → C₃

in this reaction one molecule of A₂ and one molecule of B₃ combine to from one molecule of C₃.

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2 years ago

3 examples of density synonyms

DENIUS [597]

frequency.

quantity.

thickness.

body.

closeness.

concretion.

heaviness.

solidity.

3 0

3 years ago

How many moles of each element are in one mole of Be(OH)2?

REY [17]

B. 1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen

8 0

2 years ago

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A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

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3 years ago

Calculate the theoretical yield of ammonia produced by the reaction of 100 g of H2 gas and 200g of N2 gas? 3H2(g) + N2(g)-----&g

Alex787 [66]

Moles of Hydrogen present: 100 / 2 = 50 moles

Moles of Nitrogen present: 200 / 28 = 7.14 moles

Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles

Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.

Molar ratio of Nitrogen : Ammonia = 1 : 2

Moles of ammonia = 7.14 x 2 = 14.28 moles

8 0

3 years ago

Read 2 more answers