Answer:
1.) The balanced equation is 3MgO(s) + 2Fe(s) ===> Fe2O3(s) + 3Mg(s)
2.) 20g of Fe2O3 will be produced by 20 ×120/160 = 15g of MgO
Explanation:
THE CORRECT EQUATION IS: MgO(s) + ___ Fe(s) ===> Fe2O3(s) + ___ Mg(s)
The balanced equation is: 3MgO(s) + 2Fe(s) ===> Fe2O3(s) + 3Mg(s)
molar mass of 3MgO = 3(24 + 16) = 120
Molar mass of Fe2O3 = 2 X 56 + 3 X 16
=112 +48 =160
From the equation, 120g of MgO Produced 160g of Fe2O3
Therefore 20g of Fe2O3 will be produced by 20 ×120/160 = 15g of MgO.
Answer: The corresponding value of 
for this reaction at 84.5°C is 0.00232
Explanation:
Relation of with is given by the formula:
where,
= equilibrium constant in terms of partial pressure = ?
= equilibrium constant in terms of concentration = 0.0680
R = Gas constant = 
T = temperature =
= change in number of moles of gas particles = 
Putting values in above equation, we get:
Thus the corresponding value of for this reaction at 84.5°C is 0.00232
Answer:
soil particle and method of deposition
Temperature, surface area, agitation, pressure (for gases only)
Since the Mg is in excess, therefore HCl will be fully
consumed in the reaction.
The first step is to find the amount of HCl in mol
Let N (HCl) = amount
of HCl in mol
N (HCl) = (6 mol HCL/L solution) *( 125 mL ) * (1 L/1000 mL)
= 0.75 mol of HCl
Through stoichiometry
N (H2) = 0.75 mol HCl * (1 mol H2/ 2 mol of HCl)
N(H2) = 0.375 mol H2
Since we are asked
for the number of grams of H2 (mass), we multiply this with the molar mass of hydrogen
M (H2) = 0.375 mol H2 ( 2 g H2 / 1 mol H2)
M (H2) = 0.75 g H2
