A) 405 g
B) 5 g
C) 2.67 g
D) 1 g
Correct Answer: C
Solution :
Let the volume of excess 0.8 N NaOH after the reaction with \[N{{H}_{4}}Cl\] be ?V? mL. V mL of 0.8 N NaOH \[\cong \] 40 mL of \[0.75\,N\,{{H}_{2}}S{{O}_{4}}\] \[V\times 0.8=40\,\times 0.75\] \[V=37.5\,mL\] Volume of 0.8 N NaOH consumed by \[N{{H}_{4}}Cl\] \[=100-37.5\] \[=62.5\,mL\] meq of NaOH consumed by \[N{{H}_{4}}Cl\] = meq of \[N{{H}_{4}}Cl\] used \[0.8\times 62.5=\frac{w}{53.5}\times 1000\]\[w=2.675\,g\]
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