25mL of 0.15M Pb `(NO_(3))_(2)` reacts completely with 20 mL of `Al_(2)(SO_(3))_(2)`. The molar concentration of `Al_(2)(SO_(4))_(3)` will be: `3Pb(NO

Question

25mL of 0.15M Pb `(NO_(3))_(2)` reacts completely with 20 mL of `Al_(2)(SO_(3))_(2)`. The molar concentration of `Al_(2)(SO_(4))_(3)` will be: `3Pb(NO

25mL of 0.15M Pb `(NO_(3))_(2)` reacts completely with 20 mL of `Al_(2)(SO_(3))_(2)`. The molar concentration of `Al_(2)(SO_(4))_(3)` will be:
`3Pb(NO_(3))_(2)(aq.)+Al_(2)(SO_(4))_(3)(aq.)to3PbSO_(3)(s)+2Al(NO_(3))_(3)(aq.)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - a
`(M_(1)V_(1))/(n_(1))Pb(NO_(3))_(2)=(M_(2)V_(2))/(n_(2))Al_(2)(SO_(4))_(3)`
`(0.15xx25)/(3)=(M_(2)xx20)/(1)`
`M_(2)`=0.0625

4 views Answered 2023-02-05 15:29