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adoni [48]
3 years ago
9

Considering the law of conservation of mass, if 56 g samples of solid iron (Fe) reacts with 24 g of oxygen (O2), how many grams

of Iron Oxide (FeO) are obtained?
Chemistry
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

62_grams of FeO

Explanation:

Molar mass of Fe = 55.845_g/mol

Molar mass of O2 = 15.999_g/mol

From law of conservation of mass

2 moles of Fe combines with one mole of O2 to form 2 moles of FeO

Number of moles of Fe and O2 present = mass/(molar mass) = 56/55.845~1

and 24/16 = 1.1.5

Therefore

1 part of Fe will react with 0.5 part of one mole of O2

which is 54_g of Fe reacts with 8_g to form 54+8 = 62_grams of FeO

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Question 3 0
Tanya [424]

Answer:

Option D. KBr < KCl < NaCl

Explanation:

We'll begin by calculating the number of mole of each sample.

This can be obtained as follow:

For NaCl:

Mass = 1 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 1/58.5

Mole of NaCl = 0.0171 mole

For Kbr:

Mass = 1 g

Molar mass of KBr = 39 + 80 = 119 g/mol

Mole of KBr =?

Mole = mass /Molar mass

Mole of KBr = 1/119

Mole of KBr = 0.0084 mole

For KCl:

Mass = 1 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1/74.5

Mole of KCl = 0.0134 mole

Summary

Sample >>>>>>>> Number of mole

NaCl >>>>>>>>>> 0.0171

KBr >>>>>>>>>>> 0.0084

KCl >>>>>>>>>>> 0.0134

Arranging the number of mole of the sampl in increasing order, we have:

KBr < KCl < NaCl

5 0
3 years ago
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