this is a dumb question, i am performing the following reaction: 0.005mol H2Salen + 0.022mol NaBH4 ---> H2Salan This is a reduction reaction, and H2Salan has a single hydrogen attached to each of its two nitrogens, thats the only difference. i believe i need 2 equivalents of NaBH4 as it is assumed that only one hydrogen bond reacts. so, my question is, since i need 2 equivalents, do i double the mass of NaBH4 used, thus adding 0.044mol NaBH4? which then confuses me on the theoretical yield.

Molecular weight of H2 Salen is 268g/mol

Molecular weight of NaBH4 is 37.83g/mol

Molecular weight of H2 Salan product is i believe just 270 due to the additional 2 hydrogen. so, H2 salen is the limiting reactant i think

1 H2Salen + 2NaBH4 ---> 1 H2Salan

.005mol X 268g/mol = 1.34g 0.022mol X 37.83g/mol = 0.83g but 2 equivalents so 1.66g thus 0.044mol used?

Theoretical Yield: 0.005mol X H2 Salen (270g/mol) = 1.35g is this correct?