Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

For the following reaction, the mass of water produced from

Question:

For the following reaction, the mass of water produced from $445 \mathrm{~g}$ of $\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}$ is :

$2 \mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}(\mathrm{~s})+163 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 114 \mathrm{CO}_{2}(\mathrm{~g})+110 \mathrm{H}_{2} \mathrm{OP}(1)$

  1. $495 \mathrm{~g}$

  2. $490 \mathrm{~g}$

  3. $890 \mathrm{~g}$

  4. $445 \mathrm{~g}$


Correct Option: 1

Solution:

moles of $\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}(\mathrm{~s})=\frac{445}{890}=0.5$ moles

$2 \mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}(\mathrm{~s})+163 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 114 \mathrm{CO}_{2}(\mathrm{~g})+110 \mathrm{H}_{2} \mathrm{O}(l)$

$\mathrm{n}_{\mathrm{H}_{2} \mathrm{O}}=\frac{110}{4}=\frac{55}{2}$

$m_{H_{2} \mathrm{O}}=\frac{55}{2} \times 18$

$=495 \mathrm{gm}$

Leave a comment

None
Free Study Material