3KOH + H3PO4---->K3PO4 +3H2O
a) Ta có
n\(_{H3PO4}=0,2.2=0,4\left(mol\right)\)
Theo pthh
n\(_{KOH}=3n_{H3PO4}=1,2\left(mol\right)\)
m\(_{ddKOH}=\frac{1,2.55.100}{20}=330\left(g\right)\)
b) m\(_{H3PO4}=200.1,2=240\left(g\right)\)
Theo pthh
n\(_{K3PO4}=n_{H3PO4}=0,4\left(mol\right)\)
C%=\(\frac{0,4.173}{240+330}.100\%=12,14\%\)
Ta có : nH3PO4 = 2.0,2 = 0,4 (mol)
PTHH : H3PO4+ 3KOH--->K3PO4+3H2O
=>nKOH = 1,2 (mol)
=>mKOH = 330(g)
mH3PO4 = 240(g)
=>nK3PO4 = nH3PO4 = 0,4 (mol)
=>C% = 12,14%