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3 years ago

For the balanced equation shown below, how many moles of Al will react with 98.7 moles of CuSO4

Chemistry

1 answer:

xz_007 [3.2K]3 years ago

6 0

Answer:

65.8 mol

Explanation:

Step 1: Write the balanced equation for the single displacement reaction

2 Al + 3 CuSO₄ → Al₂(SO₄)₃ + 3 Cu

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of Al to CuSO₄ is 2:3.

Step 3: Calculate the number of moles of Al that will react with 98.7 moles of CuSO₄

98.7 mol CuSO₄ × 2 mol Al/3 mol CuSO₄ = 65.8 mol Al

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odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add

Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

Explanation :

First we have to calculate the moles of $H_2SO_4$ .

$\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}$

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Now put all the given values in the above expression, we get:

$\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol$

The moles of $H_2SO_4$ is, $6.17\times 10^4mol$

Now we have to calculate the moles of $Na_2CO_3$

The balanced neutralization reaction is:

$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3$

From the balanced chemical reaction we conclude that,

As, 1 mole of $H_2SO_4$ neutralizes 1 mole of $Na_2CO_3$

So, $6.17\times 10^4mol$ of $H_2SO_4$ neutralizes

Now we have to calculate the mass of $Na_2CO_3$

$\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3$

Molar mass of $Na_2CO_3$ = 106 g/mole

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Thus, the mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

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