Question

ASAP (:

Need help with these 3 questions!!!!

1. what mass of H2 would be needed to produce 208kg of methanol?

2. How many grams do 1.3 x 10^21 atoms of sodium weigh?

3. Using the average atomic mass, calculate the number of atoms present in 1.2 mL of liquid mercury. (Density= 13.6g/mL)

Answer 1

1)
`CO (g) + 2H_(2)(g) ---> CH_(3)OH (g)`

Moles of methanol =
`208 kg CH_(3)OH * (1000 g)/(1 kg) * (1 mol CH_(3)OH)/(32.04 g CH_(3)OH) = 6492 mol CH_(3)OH`

Mass of
`H_(2)` that would produce 208 kg methanol =

`6492 mol CH_(3)OH * (2 mol H_(2))/(1 mol CH_(3)OH) * (2.02 g H_(2))/(1 mol H_(2)) * (1 kg)/(1000 g)= 26.2 kg H_(2)`

2) Mass of sodium =
`1.3 * 10^(21) atoms Na * (1 mol Na)/(6.022* 10^(23)atoms Na) * (22.99 g Na)/(1 mol Na) = 0.0496 g Na`

3) Mass of mercury =
`1.2 mL * 13.6 (g)/(mL) = 16.32 g`

Number of atoms of mercury =
`16.32 g Hg * (1 mol Hg)/(200.59 g Hg) * (6.022* 10^(23) atoms Hg)/(1 mol Hg) = 4.9 * 10^(22) atoms Hg`