The 156ml of 0.1 M HCl required to react completely with a mixture of 0.53g `Na_(2)CO_(3)` and `NaHCO_(3)` only. What was the mass of `NaHCO_(3)` in original mixture?
`Na_(2)CO_(3)+2HClrarr2NaCl+H_(2)CO_(3)`
`NaHCO_(3)+HClrarrNaCl+H_(2)CO_(3)`

Options

A. 470.4 mg
B. 510.5 mg
C. 708.5 mg
D. 900 mg

Answer

A

Explanation

`(156xx0.1)/(1000)=(0.53)/(106)xx2+(y)/(84)`
`y=[(156xx0.1)/(1000)-0.01]xx84=0.4704 "g or 470.4 mg"`