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vovangra [49]
3 years ago
15

A beaker holds 962 g of a brine solution that is 6.20 percent salt. If 123g of water are evaporated from the beaker, how much sa

lt must be added to have an 8.60 percent brine solution? How many grams of the 8.6% brine solution will be produced?
Chemistry
1 answer:
denpristay [2]3 years ago
8 0

Answer:

13.687 grams of salt should be added

The total grams of 8.6% of brine solution produced is 852.687g

Explanation:

Solution mass= 962g

Salt= 6.2%

Water = 93.8%

962 gram of water is made up of:

902.356g ( due to vaporization which reduces mass)

= 902.356 - 123

= 779. 356g of water

59.644g of salt.

If we add x gram of salt for making brine solution up to 8.6%

=(59. 644g + x)g.of salt

% salt = Mass of Salt / Total mass of solution

= 0.086= 59.644 + x / 779.356 + 59.644 + x

= 59.644 + x / 839 + x

x= 13.687 g of salt

Grams of 8.6% brine solution will be:

Gram of water + total gram of salt added to form 8.6% brine solution.

= 779.356g +59.644g + 13.687g

= 852.687g

The total grams of 8.6% of brine solution produced is 852.687g

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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

5 0
3 years ago
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