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3 years ago

How many grams of sucrose are in 1.80 of a 0.718 sucrose solution

Chemistry

1 answer:

Ronch [10]3 years ago

8 0

Answer:

445 g

Explanation:

Data Given:

Volume of sucrose solution = 1.8 L

Molarity of sucrose solution = 0.718 M

mass of solute = ?

Solution:

First we will find mole of solute in a Molar Solution of sucrose

Formula used

Molarity = No. of moles of solute / of solution in L

where

solute is sucrose

Put Values in above equation

0.718 M = No. of moles of sucrose / 1.8 L

Rearrange the equation as we have to find moles of sucrose

No. of moles of sucrose = 0.718 x 1.8

No. of moles of sucrose = 1.3 mole

Now,

we have to find mass of sucrose

For this formula will use

mass in grams = no. of moles x Molar mass . . . . . .(1)

Formula of Sucrose = C₁₂H₂₂O₁₁

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol

put values in equation 1

mass in grams = 1.3 mol x 342 g/mol

mass in grams = 444.6 g

Round the figure. So,

mass of C₁₂H₂₂O₁₁ (Sucrose) = 445 g

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In first place, the balanced reaction is:

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By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

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<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

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$amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}$