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ANTONII [103]
3 years ago
8

A metallurgical firm wishes to dispose of 1200 gallons of waste sulfuric acid whose molarity is 1.05 M. Before disposal, it will

be reacted with calcium hydroxide (slaked lime), which costs $0.25 per pound. Write the balanced chemical equation for this process. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) Determine the cost that the firm will incur from this use of slaked lime. Cost
Chemistry
1 answer:
Reil [10]3 years ago
3 0

Answer:

The balanced chemical equation for this process:

H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)

$194.51 is the cost that the firm will incur from this use of slaked lime.

Explanation:

The balanced chemical equation for this process

H_2SO_4(aq)+Ca(OH)_2(s)\rightarrow 2H_2O(l)+CaSO_4(aq)

Moles of sulfuric acid = n

Volume of sulfuric acid disposed = V = 1200 gallons = 3.785 × 1200 L = 4,542 L

1 gallon = 3.785 Liter

Morality of the sulfuric acid = M = 1.05 m

Molarity(M)=\frac{n}{V(L)}

n=m\times V=1.05 M\times 4,542 L=4,769.1 mol

According to reaction, 1 mol of sulfuric acid reacts with 1 mole of calcium hydroxide.Then 4,769.1 moles of sulfuric acid will recat with ;

\frac{1}{1}\times 4,769.1 mol=4,769.1 mol of calcium hydroxide

Mass of 4,769.1 moles of calcium hydroxide:

4,769.1 mol  74 g/mol = 352,913.4 g

= \frac{352,913.4 }{453.6} pounds =778.03 pounds

(1 pound = 453.6 grams)

Cost of 1 pound of slaked lime  = $0.25

Cost of 778.03 pounds of slaked lime  = $0.25 × 778.03 = $194.51

$194.51 is the cost that the firm will incur from this use of slaked lime.

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Since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.

<h3>What is concentration?</h3>

The term concentration has to do with the amount of substance in solution. The concentration can be measured in several units. Generally, concentration is expressed in molarity, molality, mass concentration units or percentage.

Now we are asked to find the amount concentration of calcium ions and acetate ions in a 0.80 mol/L solution of calcium acetate. The formula of calcium acetate is Ca(CH3COO)2.

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It then follows that since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.

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