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I believe it would be 15.5 g.
I am a beginner in Chemistry so please do not blame me if my answer is wrong.
Here's my solution:
Nitrate Ion is NO-3. Nitrate , Nitrogen, has an Atomic Mass of 14. Oxygen has 16, and since there's 3 of it, 16 x 3 =48
So we add them up to get their atomic mass : 48 + 14 = 62
From there I guess we should just times there all together.
DO CORRECT ME IF I AM WRONG!
0.25 mol x 62 = 15.5 g.
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∙ 11y ago
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The only information you've given is you want to know moles and how many Liters there are. To calculate, you need to know what you are preparing(i.e., N2, AgNO3, etc.), and it's Molarity(M). Here's an example: How many moles of silver nitrate are needed to prepare 250mL of standard 0.100M silver nitrate solution? Note: M = moles of solute(stuff) / volume of solution(L) = # moles / L So, ?/250mL=.1M First we use dimensional analysis to convert mL to L. 250mL x 10-3 L / 1 mL = .25L Since we know M=moles/L, we can take the Molarity and put it in moles/L form. .10M = .10 mol/L Finally, .10 mol/L x .25L = .025 mol AgNO3 We can make a general assumption for your question and say the M is standard .100M. .10M = .10 mol/L .10 mol/L x 48L = 4.8 moles
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