How many grams of Fe2+ are present in 3.84 g of iron(II) oxide? How many grams o
ID: 636827 • Letter: H
Question
How many grams of Fe2+ are present in 3.84 g of iron(II) oxide? How many grams of Fe2+ are present in 3.84 g of iron(II) oxide? How many grams of Fe2+ are present in 3.84 g of iron(II) oxide?2/ online teachin × | + agenow.com/ilrn/takeAssignment/takeCovalentActivity.do?locator assignmer Review Topics ons Use the References to access importan 1. How many grams of Fe are present in 3.84 grams of irondII) oxide? grams Fe 2. How many grams of iron() oxide contain 4.66 grams of Fetn grams iron(II) oxide. Submit Answer Retry Entire Group 8 more group attempts remaining
Explanation / Answer
1)
Molar mass of FeO,
MM = 1*MM(Fe) + 1*MM(O)
= 1*55.85 + 1*16.0
= 71.85 g/mol
mass(FeO)= 3.84 g
use:
number of mol of FeO,
n = mass of FeO/molar mass of FeO
=(3.84 g)/(71.85 g/mol)
= 5.344*10^-2 mol
This is number of moles of FeO
one mole of FeO contains 1 moles of Fe
use:
number of moles of Fe = 1 * number of moles of FeO
= 1 * 5.344*10^-2
= 5.344*10^-2
Molar mass of Fe = 55.85 g/mol
use:
mass of Fe,
m = number of mol * molar mass
= 5.344*10^-2 mol * 55.85 g/mol
= 2.985 g
Answer: 2.99 g
2)
Molar mass of Fe = 55.85 g/mol
mass(Fe)= 4.66 g
use:
number of mol of Fe,
n = mass of Fe/molar mass of Fe
=(4.66 g)/(55.85 g/mol)
= 8.344*10^-2 mol
This is number of moles of Fe
one mole of FeO contains 1 moles of Fe
use:
number of moles of FeO = number of moles of FeO / 1
= 8.344*10^-2 / 1
= 8.344*10^-2
Molar mass of FeO,
MM = 1*MM(Fe) + 1*MM(O)
= 1*55.85 + 1*16.0
= 71.85 g/mol
use:
mass of FeO,
m = number of mol * molar mass
= 8.344*10^-2 mol * 71.85 g/mol
= 6.00 g
Answer: 6.00 g
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