According to the reaction equation:
H2S + 2NaOH = Na2S + 2 H2O
1-first we need to get moles of H2S:
moles H2S = mass / molar mass
moles H2S = 1.5 g /34.082 g/mol=0.044 moles
2- then we need to get moles of NaOH:
moles of NaOH = mass / molar mass
= 2 g / 40 g/mol=0.05 g
when NaOH is the limiting reactant 0.0440 x 2 = 0.0880 moles needed
- and moles Na2S = 0.0500/2 = 0.0250
mass Na2S = moles Na2S / molar mass
0.0250 x 78.0456 g/mol = 1.95 g
3- assuming 92.0 % yield:
1.95 x 92.0/100 = 1.80 g