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lubasha [3.4K]
3 years ago
15

A health clinic uses a solution of bleach to sterilize petri dishes in which cultures are grown. the sterilization tank contains

90 gal of a solution of 4% ordinary household bleach mixed with pure distilled water. new research indicates that the concentration of bleach should be 8% for complete sterilization. how much of the solution should be drained and replaced with bleach to increase the bleach content to the recommended level?
Chemistry
1 answer:
Dahasolnce [82]3 years ago
8 0
<span>3.75 gallons need to be drained and replaced with bleach. Change the problem to "What amounts of a 100% solution and a 4% solution is needed to make 90 gallons of a 8% solution?" Given that, we'll use the following values. x = amount of 4% solution. 90-x = amount of a 100% solution. The equation to solve then becomes. 0.04 x + (90-x) = 0.08 * 90 0.04 x + 90 - x = 7.2 Add x to both sides 0.04x + 90 = 7.2 + x Subtract 0.04x from both sides 90 = 7.2 + 0.96x Subtract 7.2 from both sides 82.8 = 0.96x Divide both sides by 0.96 86.25 = x So you now know that you need 86.25 gallons of the original 4% solution and (90-86.25) = 3.75 gallons of the bleach to make the desired 90 gallons. So simply drain 3.75 gallons from the tank and replace with bleach.</span>
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45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

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2 atoms of Al required 3 molecules of Cl₂.

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30 atoms of Al will require = \frac{30 * 3}{2}\\ = 45 molecules of Cl₂.

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If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
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Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

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