In the human body, the toxic compound hydrogen cyanide `HCN` is neutralized by the acid, `H_(2)S_(2)O_(3)`, according to the reaction: `HCN+H_(2)S_(2)O_(3)->HCNS+H_(2)SO(3)` If 1.00 mg of `H_(2)S_(2)O_(3)`, is available in the body, will this be enough to neutralize 2.00 mg of HCN swallowed by a person?
1 Answer | Add Yours
The reaction is balanced, so we don't have to change the ratios of each compound to each other. `H_(2)S_(2)O_(3)` has a molecular mass of 114g/mol. So 1.00 mg, or .001g would contain `(1g)/1000 xx (1 mol)/(114g)=8.77xx10^-6 mol` of `H_(2)S_(2)O_(3)`
`HCN` has a molecular mass of 27g/mol. So 2.00 mg or .002g would contain `(2g)/1000 xx (1mol)/(27g)=7.41xx10^-5 mol` of `HCN`
The ratio of reactants in the equation is 1:1 so we must have at least as many mols of `H_(2)S_(2)O_(3)` as we have mols of `HCN` Since we have nearly 10 times more mols of `HCN` than we have of `H_(2)S_(2)O_(3)` we do not have enough to neutralize all the poison.
We’ve answered 318,916 questions. We can answer yours, too.Ask a question