P.V= n.R.T
n= P.V/R.T; n= (700/760. 1) / (0,082 x (273+20)) = 0,0383 moles de NH3
1 mol de NH4Cl da 1 mol de NH3.
por tanto hace falta 0,0383 moles de NH4Cl
g= moles x Mm = 0,0383 x 53,5 = 2,05 g de NH4Cl
14 + 4 + 35,5 = 53,5 g/mol
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P.V= n.R.T
n= P.V/R.T; n= (700/760. 1) / (0,082 x (273+20)) = 0,0383 moles de NH3
1 mol de NH4Cl da 1 mol de NH3.
por tanto hace falta 0,0383 moles de NH4Cl
g= moles x Mm = 0,0383 x 53,5 = 2,05 g de NH4Cl
14 + 4 + 35,5 = 53,5 g/mol