Answer:
70.3824 grams of carbondioxide is produced.
Explanation:
![C_6H_{12}O_6(aq)+6O_2(g)⟶6CO_2(g)+6H_2O(l)](https://tex.z-dn.net/?f=C_6H_%7B12%7DO_6%28aq%29%2B6O_2%28g%29%E2%9F%B66CO_2%28g%29%2B6H_2O%28l%29)
Moles of glucose = ![\frac{48.0 g}{180 g/mol}=0.2666 mol](https://tex.z-dn.net/?f=%5Cfrac%7B48.0%20g%7D%7B180%20g%2Fmol%7D%3D0.2666%20mol)
According to reaction 1 mole of glucose reacts with 6 mole of oxygen gas.
Then 0.2666 mol of gluocse will reactwith :
of oxygen
Mass of 1.5996 moles of oxygen gas:
![1.5996 mol\times 32 g/mol = 51.1872 g](https://tex.z-dn.net/?f=1.5996%20mol%5Ctimes%2032%20g%2Fmol%20%3D%2051.1872%20g)
51.1872 grams of oxygen are required to convert 48.0 grams of glucose to
and
.
According to reaction, 1 mol, of glucose gives 6 moles of carbon dioxide.
Then 0.2666 moles of glucose will give:
of carbon dioxide
Mass of 1.5996 moles of carbon dioxide gas:
![1.5996 mol\times 44 g/mol = 70.3824 g](https://tex.z-dn.net/?f=1.5996%20mol%5Ctimes%2044%20g%2Fmol%20%3D%2070.3824%20g)
70.3824 grams of carbondioxide is produced.