A) 0.01g of hydrogen
B) 0.085g of \[N{{H}_{3}}\]
C) 320 mg of gaseous \[S{{O}_{2}}\]
D) all of the above
Correct Answer: D
Solution :
\[0.22g\,C{{O}_{2}}=\frac{0.22}{44}\text{mol}=5\times {{10}^{-3}}\text{mol}\] \[=5\times {{10}^{-3}}\times 22400\,cc=110\,cc\] (a) \[0.01g\,{{H}_{2}}=\frac{0.01}{2}\text{mol}=5\times {{10}^{-3}}\text{mol}\] (b)\[0.085g\,\text{of}\,N{{H}_{3}}=\frac{0.085}{17}\text{mol}=5\times {{10}^{-3}}\text{mol}\] (c)\[320\,\text{mg}\,S{{O}_{2}}=\frac{0.320}{64}\text{mol}=5\times {{10}^{-3}}\text{mol}\]
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