The formula of Iron(III) oxide is Fe2O3
In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100
= [(moles of iron * Mr of iron) / (moles of Iron * Mr of Iron + moles of Oxygen * Mr of Oxygen)] * 100
= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
= (112 / 160) * 100
= 70%
Thus, in a 100g sample, the weight of iron will be:
100 * 70%
= 70 grams
Iron is left in the filter and salt solution (salt and water) passes into the cup.
Hope it helps
E
Water is the only polar compound so I guess that must be odd one out. I’m sorry I’m of no actual help.
Answer:
Explanation:
10.11 g sample of 
contains 22.34% 
by mass
According to the law of constant composition the if one sample of has 22.34% of by mass then any other sample of will have the same percentage of the amount of .
For a sample of 7.09 g we have
The mass of sodium in the required sample is .
