Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
The answer is B. The density, or more precisely, the volumetric mass density, of a substance is its mass per unit volume.
Answer:
2.67 × 10⁻²
Explanation:
Equation for the reaction is expressed as:
CaCrO₄(s) ⇄ Ca₂⁺(aq) + CrO₂⁻⁴(aq)
Given that:
Kc=7.1×10⁻⁴
Kc= ![[Ca^{2+}][CrO^{2-}_4]](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5BCrO%5E%7B2-%7D_4%5D)
Kc= [x][x]
Kc= [x²]
7.1×10⁻⁴ = [x²]
x = 
x = 0.0267
x = 
Moles of ammonium sulfide = 5.80 mol
The formula of ammonium sulfide is (NH₄)₂S
So each molecule of ammonium sulfide has (4 x 2) or 8 atoms of H
One mole of ammonium sulfide has 8 moles of H
5.80 mol of ammonium sulfide has (8 x 5.8) or 46.4 moles of H
As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ atoms
46.4 moles of H x (6.022 x 10²³ atoms/ 1 mole of H)
= 2.8 x 10²⁵ H atoms
Therefore, 2.8 x 10²⁵ H atoms are in 5.80 mol of ammonium sulfide.
