Question

What mass of oxygen (O2) forms in a reaction that forms 15.90 g C6H12O6? (Molar mass of O2 = 32.00 g/mol; molar mass of C6H12O6 = 180.18 g/mol) 15.90 g C6H12O6 = g O2

Answer 1

The answer is: the mass of oxygen is 16.95 grams.

The overall balanced photosynthesis reaction:  

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.  

m(C₆H₁₂O₆) = 15.90 g; mass of glucose.

n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).

n(C₆H₁₂O₆) = 15.9 g ÷ 180.18 g/mol.

n(C₆H₁₂O₆) = 0.088 mol; amount of glucose.

From chemical reaction: n(C₆H₁₂O₆) : n(O₂) = 1 : 6.

n(O₂) = 6 · 0.088 mol.

n(O₂) = 0.53 mol; amount of oxygen.

m(O₂) = 0.53 mol · 32.00 g/mol.

m(O₂) = 16.95 g; mass of oxygen.

Answer 2

Answer : The mass of $O_2$ forms are 16.94 grams.

Explanation : Given,

Mass of $C_6H_{12}O_6$ = 15.90 g

Molar mass of $C_6H_{12}O_6$ = 180.18 g/mole

Molar mass of $O_2$ = 32.00 g/mole

The balanced chemical reaction will be:

$6CO_2(g)+6H_2O(g)\rightarrow C_6H_{12}O_6(s)+6O_2(g)$

First we have to calculate the moles of $C_6H_{12}O_6$

$\text{ Moles of }C_6H_{12}O_6=\frac{\text{ Mass of }C_6H_{12}O_6}{\text{ Molar mass of }C_6H_{12}O_6}=\frac{15.90g}{180.18g/mole}=0.08824moles$

Now we have to calculate the moles of $O_2$

As, 1 mole of $C_6H_{12}O_6$ produced when 6 moles of oxygen produced

So, 0.08824 mole of $C_6H_{12}O_6$ produced when $0.08824\times 6=0.5294$ moles of oxygen produced

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(0.5294moles)\times (32.00g/mole)=16.94g$

Therefore, the mass of $O_2$ forms are 16.94 grams.