Answer:
1.98*10^18 particles of H2
Step-by-step explanation:
The balanced equation is
CO(g) + 2H2 (g) -> CH3OH(l)
The number of moles of each component is calculated as follows:
moles = mass/molecular weight
where mass is in grams and molecular weight in grams/mol
Then:
moles of CO = (1.5x10^-6)/(28) = 5.35*10^-8 mol
moles of H2 = (6.8x10^-6)/(2) = 3.4*10^-6 mol
From the balanced equation we now that 2 moles of H2 will react with 1 mole of CO. Then, if 5.35*10^-8 mol of CO2 reacts, the following proportion must be satisfied:
2 moles of H2/x moles of H2 = 1 mole of CO/5.35*10^-8 mol of CO2
x = 5.35*10^-8/1 * 2 = 1.07*10^-7 moles of H2
But 3.4*10^-6 mol of H2 entered to the vessel, so 3.4*10^-6 - 1.07*10^-7 = 3.293*10^-6 moles of H2 will left after the completion of the reaction (all CO reacted). Using Avogadro's number we can calculate the number of particles would remain in the reaction vessel:
3.293*10^-6 mole of H2 * 6.022×10^23 particles/mole = 1.98*10^18 particles of H2