What mass of methanol (CH3OH) is produced when 140.1 g of carbon monoxide reacts with 12.12 g of hydrogen?
co(g)+2h2(g)--->ch3oh(/)
a96.12g
b160.2g
c192.2g
d6.06.g

Explanation:This is a limiting reactant problem.  One way to deal with limiting reactant problems is to compute the amount of product using each of the reactants.  The theoretical yield will be the lesser of the two and that will also indicate the limiting reactant.

CO(g) + 2H2(g) --catalysts--> CH3OH(g)

140.1g..12.12g ?g

140.1g CO x (1 mol CO / 28.0g CO) x (1 mol CH3OH / 1 mol CO) x (32.0g CH3OH / 1 mol CH3OH) = 160.1g CH3OH

12.12g H2 x (1 mol H2 / 2.02g H2) x (1 mol CH3OH / 2 mol H2) x (32.0g CH3OH / 1 mol CH3OH) = 96.00g CH3OH

96.00g CH3OH will be produced and hydrogen is the limiting reactant.

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explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.