Can someone help me with this problem

The equation for the combustion of glucose is

C6H12O6(s) + 6O2(g) ® 6CO2(g) + 6H2O(g)

How many grams of CO2 will be produced when 9.554 g of glucose is burned?

i'm really not sure with this problem, and it's suppose to be easy i think. help me please, i need to do more problems like this and this will be the model.

In the equation you got this proportion
1 : 6 = 6 : 6

This means that when you burn 1 mole of glucose with 6 mole of oxygen you got ONLY 6 mole of CO2 and 6 of water.
1 mole of glucose equals 180 g and 1 mole of CO2 equals 44 g (consequently 6 mole of CO2 equals 264 g).

Now you can do the proportion. If 1 mole of glucose (with 6 mole of oxygen) produces 6 mole of CO2 consequently:
180g (glucose) : 264g (CO2) = 9.554g (glucose) : Xg (CO2)
X = 9.554 x 264 / 180 = 14.012 g (CO2)

You must use stoichiometry to solve this problem. Looking at your equation, you can see that for every one mole of glucose that is burned, 6 moles of CO2 are produced. Using this info, you can set up your equation as follows:

9.554g x (1/molar mass of glucose) x (6 moles of CO2/ 1 mole of glucose) x (1/ 44g/mol CO2)

The final answer should be 0.007g CO2

Now I'll explain the equation. You first want to convert glucose into moles by dividing it by its molar mass of 180g/mol. Now that you have moles of glucose, you multiply by six according to your equation. For every 1 mole of glucose burned, six moles of CO2 are produced. After multiplying by this number, divide by the molar mass of CO2 which is 44g/mol to get the mass of CO2 produced.

Molecular weight of C6H12O6 = 12x6+12x1+16x6 = 180g
ie,1 mole of C6H12O6 = 180g
1 mole of glucose gives 6 moles of CO2(given)
(1x180)g of glucose -------> (6x 44)g of CO2
ie,180 g of glucose -------->264g of CO2
Therefore,1g of glucose gives------>264/180 = 1.467g of CO2
so,9.554g of glucose gives-------->(1.467x9.554)g of CO2
=14.02g of CO2

This is a fundamental stoichiometry problem.

9.554 g C6H12O6 * ( 1 mole /180 g C6H12O6) * (6 mole CO2 / 1 mole) * ( 44.0 g CO2 / 1 mole) = you do the math.

Note that the answer will have 4 sig figs since the initial number, 9.554, had 4 sig fig.