Use PV = nRT and solve for mols NH3. I get about 0.016 but you need to do it more accurately.
2NH3 + H2SO4 => (NH4)2SO4
N H2SO4 = # equivalents/L
#equivalents NH3 = 0.016
2 mols NH3 = 1 mol H2SO4
or 1 equivalent NH3 = 1/2 mol H2SO4
N H2SO4 = approx 0.016/2/0.134 = ?
2 litre of NH3 at 30°C and at 0.20 atmosphere is neutralized by 134 mL of a solution of H2SO4. Calculate normality of H2SO4
1 answer
1 answer