I don't have a calculator with me right now, but that mass would be 1200 grams. Divide the given amount of grams by the molar mass of NH3, which is 17.031g/mol. (Nitrogen + 3(hydrogen)). Again, sorry I didn't have a calculator. But all you would need to do is divide 1200 by 17.031. If you need to use sig figs, your answer should have 2 because the 1.2 x 10^3 limits your amount of sig figs.

6 0

3 years ago

Our good friend and pseudo scientist, homer simpson, attempts to analyze 300 mg of an unknown compound containing only c and h b

iragen [17]

$C_{x}H_{y} + O_{2} --\ \textgreater \ x CO_{2} + \frac{y}{2} H{2}O

540mg H_{2}O* \frac{1 mmol H_{2}O}{18 mg H_{2}O} =30 mmol H_{2}O

\\ \\30 mmolH_{2}O -have - 60 mmol H
\\ \\ 60 mmol H* \frac{1mg H}{1 mmol H} =60 mg H
\\ \\300mg C_{x}H_{y} - 60 mg H= 240 mg C
\\ \\ 240 mgC* \frac{1mmol}{12mg} =20 mmol C

20 mmol C: 60 mmol H=1 mol C : 3 mol H

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7 0

3 years ago