We are given the chemical reaction and the amount of reactant used for the process. We use these data together to obtain what is asked. We do as as follows:
0.882 mol H2O2 ( 1 mol O2 / 2 mol H2O2 ) = 0.441 mol O2 produced
Hope this answers the question.
Cl2 is nonpolar so it has to be only London dispersion force (LDF)
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y
3y = 1.2
y = 0,4M
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4
C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄
mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol
164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄
