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4 years ago

How many moles of co2 are produced when 35.0 mol c2h2 react completely?

Chemistry

2 answers:

Setler [38]4 years ago

7 0

The balanced chemical reaction will be:

C2H2 + 3/2O2 = 2CO2 +H2O

We are given the amount of C2H2 to be burned. This will be our starting point.

35.0 mol C2H2 ( 2 mol CO2/1 mol C2H2) = 70.0 mol O2

Therefore, the CO2 produced is 70.0 moles.

yulyashka [42]4 years ago

3 0

Answer: The moles of $CO_2$ produced will be 70 moles.

Explanation:

The chemical reaction for the combustion of ethyne follows the equation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of ethyne produces 4 moles of carbon dioxide.

So, 35 moles of ethyne will produce = $\frac{4}{2}\times 35=70moles$ of carbon dioxide.

Hence, the moles of $CO_2$ produced will be 70 moles.

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vlada-n [284]

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The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

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$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

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$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

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