I´m pretty sure its magnesium iron silicate.
Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
It would be 6 because of the elements
Answer:
We need 226 grams of FeS
Explanation:
Step 1: Data given
Mass of FeCl2 = 326 grams
Molar mass FeCl2 = 126.75 g/mol
Step 2: The balanced equation
FeS + 2 HCl → H2S + FeCl2
Step 3: Calculate moles FeCl2
Moles FeCl2 = 326 grams / 126.75 grams
Moles FeCl2 = 2.57 moles
Step 4: Calculate moles FeS needed
For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl
For 2.57 moles FeCl2 we need 2.57 moles FeS
Step 5: Calculate mass FeS
Mass FeS = 2.57 moles * 87.92 g/mol
Mass FeS = 226 grams FeS
We need 226 grams of FeS
Answer is: 230 g.
Chemical reaction: P₄ + 5O₂ → 2P₂O₅.
m(P₄) = 100 g.
M(P₄) = 4 · 31 g/mol = 124 g/mol.
n(P₄) = m(P₄) ÷ M(P₄) = 100g ÷ 124g/mol = 0,806 mol.
From reaction: n(P₄) : n(P₂O5) = 1 : 2.
n(P₂O₅) = 1,612 mol.
m(P₂O₅) = 1,612 mol · 142g/mol = 230g.
M - molar mass.
n - amount of substance.
