The relative molecular mass of magnesium chloride is approximately 60. The molar mass is therefore 60g per mole. Therefore there is 0.42mol of formula units in 2.5 grams.
The formula mass of MgCl2 is 24.3 + 2(35.5) = 95.3Amount of MgCl2 = 2.5/95.3 = 0.0262mol
There are 0.0262 moles of formula unit in 2.5 grams of magnesium chloride.
To get the number (not in moles), multiply the amount in moles by the Avogadro's constant.
Current is defined as the quantity of charge passing a certain point each second given by:
I= Q/t ,where t is time (in seconds) and Q is the charge (in Coulombs).
91.06g of chromium metal (Cr(s)) translates to
n=m/M=91.6g/52.00g/mol= 1.762 mol
Next we must figure out how many electrons are required per reduction reaction.
Cr4+ + 4e- ----> Cr(s), therefore we need 4 electrons per chromium atom to convert it from the oxide to the metal.
#of e- = 4(1.762 mol) = 7.046 mol e-
Lastly we need to convert this number of electrons into a charge (in couloubs).
The charge of one electron is e = 1.6E-19 C, so the total charge is:
Q=(7.046 mol e-)(1.6E-19 C)(6.022E23 e-/mol)= 678900 C
t = (12.4 h)(60min/h)(60s/min)= 44640 s
I= 678900 C/44640 s = 15.2 A
Therefore the current required to produced 91.6 g of chromium metal in 12.4 hours is 15.2 A.
Take Fe(OH)2 as an example where '2' is in subscript. It means that in one formula unit there is one Fe and two hydroxyl (-OH) groups. The two outside the brackets tell us that it apply to not just the oxygen or the hydrogen, but to both of them as a hydroxyl group.
It is an example for a combustion reaction. The matchstick contains sulfur and carbon. Therefore two of the products would be sulfur dioxide and carbon dioxide.
117(g K) * [1.008(g/mol H) / 39.098(g/mol K)] = 3.02 gram hydrogen
That depends on which molecule you are referring to when you say "iron oxide". All of the following are correctly referred to as "iron oxide" - either as iron (II) oxide or iron (III) oxide
FeO
Fe3O4
Fe4O5
Fe5O6
Fe5O7
Fe25O32
Fe13O19
Fe2O3
Assuming you are looking to find the number of grams of oxygen and iron respectively required to produce 100 grams of "iron oxide" you would have to refer to the atomic weights of iron (55.845) and oxygen (~ 15.999 or 16) and then use them to find the molecular weight of the chosen form of iron oxide. From that you would calculate the number of grams required from the formula:
For FemOn
g oxygen =
100 g x 1 mole FemOn/([55.845 x m] + [15.999 x n])g FemOn x n moles O/mole FemOn x 15.999 g O/mole O.
g iron =
100 g x 1 mole FemOn/([55.845 x m] + [15.999 x n])g FemOn x m moles Fe/mole FemOn x 55.845 g Fe/mole Fe.
When sodium potassium permanganate and hydrogen peroxide interact 30% of oxygen gas is usually produced from the mixture.
Well first things first you have to balance the equation
Fe3O4-->3Fe+2O2
Then because your given mass is in kg you have to convert it to grams
100.0kgx1,000kg/(over)1g=100,000g or 1.0x105
Next because you have mass of Fe3O4 you need to get the molar mass.
(3)55.85gFe+(4)16.00gO=231.55gFe3O4
Now convert grams to moles for your final answer
1.0x105gFe3O4x1mol/(over)231.55g(mm)x3molFe/(over)1mol Fe=1295.6 mol Fe
you can recover 1295.6moles of Fe from Fe3O4
If you follow the rule of Significant Figures, 100.0kg has 4 significant figures (digits) so your answer needs to have 4 digits
1295.6=1296
you can recover 1296 moles of Fe from Fe3O4
It could be argued that the concept of stoichiometry was first formally stated by John Dalton, who stated as part of his Atomic Theory that:
Assuming you've asked recently, I can help you. just ask questions and tag it in the chem category and I'll answer them. I'll still be online for a few more hours.- Jen.
Stoichiometry is based on the fact that chemical reactions occur in ratios of moles of all substances. The problem is that there is no device that directly measures moles. We do have a device to measure mass, so molar mass is used in stoichiometric calculations to give results which are measurable.
We convert units to moles because in reactions atoms of different elements combine in simple whole-number ratios. The whole numbers are related to the number of atoms, not their weights.
As stated in the category description: "Stoichiometry is a branch of chemistry which deals with the ratios of the reactants and products involved in chemical equations."
Limiting reactants are the reactants that are used up first. And once they are used up, they stop, or limit, the reaction. So the amount of product that can be produced depends on the limiting reactant.
The other reactant, the one in excess, would predict a larger amount of product. But once we produce the amount of product predicted by the limiting reactant. The limiting reactant is used up and the reaction stops.