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3 years ago

What mass of carbon is present in 0.500 mol of sucrose?

Chemistry

2 answers:

Vilka [71]3 years ago

5 0

Answer:

There are 72 g of carbon in 0.500 mol of sucrose.

Explanation:

The sucrose molecule has the formul C₁₂H₂₂O₁₁.

Each molecule of sucrose has 12 atoms of carbon.

Each carbon atom weights 12 a.u.

Each mole of sucrose weights 342.30 g.

Each mole of carbon weights 12 g.

12 carbons weight 144 g.

Therefore, the weight percentage of carbon in each mole of sucrose is:

%C = 144/342.30 x 100%

%C = 42.1%

0.500 mol of sucrose weights 342.30 g/2 = 171.15 g.

42.1% x 171.15 g = 72 g

AlexFokin [52]3 years ago

3 0

C12H22O11
12×12×0.5=72

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Solution : Given,

Mass of $Ag_2O$ = 25.0 g

Mass of $C_{10}H_{10}N_4SO_2$ = 50.0 g

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First we have to calculate the moles of $Ag_2O$ and $C_{10}H_{10}N_4SO_2$ .

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$\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)$

From the balanced reaction we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 1 mole of $Ag_2O$

So, 0.1998 moles of $C_{10}H_{10}N_4SO_2$ react with $\frac{0.1998}{2}=0.0999$ moles of $Ag_2O$

From this we conclude that, $Ag_2O$ is an excess reagent because the given moles are greater than the required moles and $C_{10}H_{10}N_4SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgC_{10}H_9N_4SO_2$