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I'm an amateur analog photographer and want to react Nitric Acid (62%) and silver to Silver nitrate. Silver nitrate is the basis for almost all black and white photography.

I'm trying to figure out the correct ratio of silver to nitric acid. The reaction is $$\ce{3Ag(s) +4HNO3(aq) -> 3AgNO3(l) +NO(g) +2H2O (l)}$$

Can I calculate the ratio of silver to acid with the formula $3 \times (\text{molar weight of Ag}) / 4 \times (\text{Molar weight of HNO3}) \times 0.62$ ?

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  • $\begingroup$ This is just a bad idea. Molar proportion's your last concern with this reaction. $\endgroup$ – Mithoron Aug 16 at 18:16
  • $\begingroup$ I'm perfectly aware of the reaction and it's bi product. Thanks for your concern! $\endgroup$ – Glenn Bech Aug 16 at 18:55
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Every 3 moles of Ag reacts with 4 moles of Nitric acid.
Therefore the mass ratio is: $$ \frac{m(Ag)}{m(HNO3)} = \frac{3}{4}\frac{MM(Ag)}{MM(HNO3)} = 1.28 $$

MM is molar mass(g/mol)

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  • $\begingroup$ I have to multiply 4 MM(HNO3) by a factor of 0.62 to take the diluted acid into account, right? $\endgroup$ – Glenn Bech Aug 16 at 20:37
  • $\begingroup$ So you want the ratio: silver mass/volume HNO3 solution? $\endgroup$ – blu potatos Aug 16 at 21:46
  • $\begingroup$ Silver mass / mass of dilluted HNO3 preferably. $\endgroup$ – Glenn Bech Aug 17 at 11:12

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