C is the correct answer. Double displacement reaction, also called as metathesis, is a type of reaction wherein two compounds react to form new compounds. This case the cations and the anions of the reactants replace each other forming the new compounds.
14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.
Explanation:
The balanced equation for the reaction is to be known so that number of moles taking part can be known.
The balanced chemical equation is
4Fe + 3
⇒ 2 
From the given weight of iron to be used for the production of , number of moles of Fe taking part in the reaction can be known by the formula:
Number of moles= mass ÷ Atomic mass of one mole of the element.
(Atomic weight of Fe is 55.845 gm/mole)
Putting the values in equation
Number of moles = 10 gm ÷ 55.845 gm/mole
= 0.179 moles
Applying the stoichiometry concept
4 moles of Fe gives 2 Moles of Fe2O3
0.179 moles will produce x moles of Fe2O3
So, 2÷ 4 = x ÷ 0.179
2/4 = x/ 0.179
2 × 0.179 = 4x
2 × 0.179 / 4 = x
x = 0.0895 moles
So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.
Now the formula used above will give the weight of Fe2O3
weight = atomic weight × number of moles
= 159.69 grams × 0.0895
= 14.292 grams of Fe2O3 formed.
Answer: It becomes the uncombined element in the product.
Explanation:
The reaction between Zn and HCl is a single displacement reaction according to equation below
Zn + 2HCl —> ZnCl2 + H2
Zn displaces H2 from acid and in the product, hydrogen became the uncombined element.
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
