How many molecules of (CH3)2NH are needed to react completely with 2.00mg of H2C2O4? Help i don't understand this process!

See your post above.

To answer this question, you need to use the concept of stoichiometry. Stoichiometry relates the balanced chemical equation to determine the number of molecules involved in a reaction.

First, we need to write the balanced chemical equation for the reaction between (CH3)2NH and H2C2O4. Let's assume the balanced equation is:

2(CH3)2NH + H2C2O4 → 2(CH3)2NCOOH + H2O

Next, convert the given mass of H2C2O4 to moles. We can do this by using the molecular weight of H2C2O4, which is 90.03 g/mol.

Mass of H2C2O4 = 2.00 mg = 0.002 g
Moles of H2C2O4 = (0.002 g) / (90.03 g/mol) = 2.22 x 10^-5 mol

According to the balanced equation, the stoichiometric ratio between H2C2O4 and (CH3)2NH is 1:2. This means that 1 mole of H2C2O4 reacts with 2 moles of (CH3)2NH.

Using this information, we can calculate the moles of (CH3)2NH required to react completely:

Moles of (CH3)2NH = (2.22 x 10^-5 mol H2C2O4) x (2 mol (CH3)2NH / 1 mol H2C2O4) = 4.44 x 10^-5 mol (CH3)2NH

Finally, we can convert the moles of (CH3)2NH to molecules by using Avogadro's number. Avogadro's number is 6.022 x 10^23 molecules/mol.

Number of molecules of (CH3)2NH = (4.44 x 10^-5 mol) x (6.022 x 10^23 molecules/mol) = 2.67 x 10^19 molecules

Therefore, approximately 2.67 x 10^19 molecules of (CH3)2NH are needed to completely react with 2.00 mg of H2C2O4.