Answer: m = 50 g ZnSO4
Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.
0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn
= 0.311 moles ZnSO4
0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4
= 50 ZnSO4
First. let's write the reaction formula: HBr +LiOH ----> LiBr + H₂O
let's get the moles of LiOH first
moles= Molarity x Liters
moles= 0.253 M x 0.01673 Liter= 0.00423 moles LiOH
using the balanced equation, you can see that 1 mol LiOH is equal to 1 mol HBr. so:
0.00423 mol LiOH = 0.00423 mol HBr
now let's find the concentration
molarity= mol/ Liters
0.00423 mol/ 0.01000 Liters= 0.423 M
Answer: Oxidation number of chlorine in potassium chlorate...
so, oxidation state of chlorine in potassium chlorate is +1. and yea!!
Explanation: hope this help
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