Answer:
FeCl₃
Explanation:
4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂
Given => 7moles 9moles
A simple way to determine which reagent is the limiting reactant is to convert all given data to moles then divide by the respective coefficients of the balanced equation. The smaller value will be the limiting reactant.
4FeCl₃ + 3O₂ => 2Fe₂O₃+ 6Cl₂
Given => 7/4 = 1.75* 9/3 = 3
*Smaller value => FeCl₃ is limiting reactant.
NOTE: However, when working problems, one must use original mole values given.
It is by looking on the dates of the paper.
Hoped this helped.
~Bob Ross®
The number of Ml of a 0.40 %w/v solution of ,nalorphine that must be injected to obtain a dose of 1.5 mg is calculated as below
since M/v% is mass of solute in grams per 100 ml
convert Mg to g
1 g = 1000 mg what about 1.5 mg =? grams
= 1.5 /1000 = 0.0015 grams
volume is therefore = 100 ( mass/ M/v%)
= 100 x( 0.0015/ 0.4) = 0.375 ML
The journey from earth to the nearest plant will take the longest
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
