Chemical quantities (AS)

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How do I work out the quantity of nitric acid needed to react with 2.45 g of MgCO₃?
 
I know the reaction involves a 1:2 molar ratio but how can I use this to find out the exact quantity of acid needed?

Igloo writes
 
You are correct: magnesium carbonate and nitric acid do react in a 1:2 molar ratio, so what we must do first is to convert the 2.45g of MgCO₃ into moles.
Since M_r[MgCO₃] = 24 + 12 + (3x16) = 84 g mol^-1
Moles of magnesium carbonate = 2.45 / 84 = 0.0292
So the quantity of nitric acid required = 2 x 0.0292 = 0.058 mol (2 s.f.)
 
Your question doesn’t make it clear whether you need the answer in moles or grams, or whether you are using dilute nitric acid, which is how this reaction would be carried out safely, in practice. If you want to find the volume of dilute nitric acid required you will need to know its concentration. I’ll leave you to complete the calculation from here.

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updated: 25 January 2006