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3 years ago

Determine the mass in grams of each nacl solution that contains 1.5g of nacl

Chemistry

1 answer:

Rama09 [41]3 years ago

7 0

Answer:- 3333 g of solution.

Some of the question part is missing here. It would be like, "Determine the mass in grams of each NaCl solution that contains 1.5 g of NaCl.

(i) 0.045% NaCl by mass

Solution:- 0.045% NaCl by mass means 0.045 g of NaCl are present in 100 g of solution. 1.5 g of NaCl would be present in how many grams of solution?

We could solve this using proportions...

(0.045/100) = (1.5/X)

0.045(X) = 1.5(100)

0.045X = 150

X = 150/0.045 = 3333

So, 1.5 g of NaCl is present in 3333 g of solution.

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What is a sign that a chemical change has occurred?

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The production of heat, light, or smoke is observed.

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If 5.65 grams of zinc metal react with 21.6 grams of silver nitrate, how many grams of silver metal can be formed and how many g

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The balanced chemical reaction is:

Zn + 2AgNO3 = Zn(NO3)2 + 2Ag

To determine the amount of the reactant left, we have to determine which is the limiting and the excess reactant. We do as follows:

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6.0 mol NaOH reacts with

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Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄

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:

NaOH: 3 moles
H₃PO₄: 1 mole
H₂O: 3 moles
Na₃PO₄: 1 mole

<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 H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?

$moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}$

moles of NaOH= 27 moles

But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.

<h3>Moles of Na₃PO₄ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?

$moles of Na_{3}P O_{4} =\frac{6 moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}$

<u><em>moles of Na₃PO₄= 2 moles</em></u>

Then, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.

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