Answer:
13.92 %
Explanation:
Mass of 
= 12.5221 g
Molar mass of = 233.43 g/mol
The formula for the calculation of moles is shown below:
Thus,
Moles of = 0.0536 moles
According to the given reaction,
1 mole of is formed from 1 mole of 
Thus,
0.0536 moles of is formed from 0.0536 moles of
Moles of = 0.0536 moles
Moles of sulfur in 1 mole = 1 mole
Moles of sulfur in 0.0536 mole = 0.0536 mole
Molar mass of sulfur = 32.065 g/mol
Mass = Moles * Molar mass = 0.0536 * 32.065 g = 1.7187 g
Mass of ore = 12.3430 g
Mass % = 
= 
= 13.92 %
6787.67458865446899675566
CO(g) +2H2--->CH3OH
2.50g H2*1mol/2g=1.25 mol H2
30.0L CO*1mol/22.4L=1.34 mol CO,
according to reaction 1 mol CO needs 2 mol H2,so 1.34 mol CO need 2.68 mol H2, so 1) limiting teactant is H2 (H)
2)1.25 mol CH3OH will be produced, 1.25 mol*32g/mol=40.0 g CH3OH
3) 1.25 mol H2 needs 0.625 g CO
1.34-0.625=0.715 g CO leftover
We will use the expression for freezing point depression ∆Tf
∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is
∆Tf = 0C - (-3°C) = 3°C
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and one Cl- ion:
KCl → K+ + Cl-
Therefore, the molality m of the solution can be calculated as
3 = 2 * 1.86 * m
m = 3 / (2 * 1.86)
m = 0.80 molal
