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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)

Respuesta :

Chlidonias

1) [tex] CO (g) + 2H_{2}(g) ---> CH_{3}OH (g) [/tex]

Moles of methanol = [tex] 208 kg CH_{3}OH * \frac{1000 g}{1 kg} * \frac{1 mol CH_{3}OH}{32.04 g CH_{3}OH} = 6492 mol CH_{3}OH [/tex]

Mass of [tex] H_{2} [/tex] that would produce 208 kg methanol =

[tex] 6492 mol CH_{3}OH * \frac{2 mol H_{2}}{1 mol CH_{3}OH} * \frac{2.02 g H_{2}}{1 mol H_{2}} * \frac{1 kg}{1000 g}= 26.2 kg H_{2} [/tex]

2) Mass of sodium = [tex] 1.3 * 10^{21} atoms Na * \frac{1 mol Na}{6.022* 10^{23}atoms Na} * \frac{22.99 g Na}{1 mol Na} = 0.0496 g Na [/tex]

3) Mass of mercury = [tex] 1.2 mL * 13.6 \frac{g}{mL} = 16.32 g [/tex]

Number of atoms of mercury = [tex] 16.32 g Hg * \frac{1 mol Hg}{200.59 g Hg} * \frac{6.022* 10^{23} atoms Hg}{1 mol Hg} = 4.9 * 10^{22} atoms Hg [/tex]

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