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3 years ago

How many moles of H2O are produced when 64.0 g C2H2 burns in oxygen

1 answer:

6 0

The balanced chemical reaction will be:

C2H2 + 5/2O2 = 2CO2 + H2O

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

64.0 g C2H2 (1 mol C2H2 / 26.02 g C2H2) (1 mol H2O/1mol C2H2) ( 18.02 g H2O/1mol H2O) = 44.32 g H2O.

Thus, the answer is 44.32 g H2O.

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Given that this reaction is exothermic, what direction will the equilibrium shift when the temperature of the reaction is decrea

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Equilibrium will shift towards the products when temperature is decreased in an exothermic reaction of the formation of ammonia.

<h3>What is an exothermic reaction?</h3>

An exothermic reaction is a reaction in which heat content of the reactants is greater than the heat content of product.

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Learn more about exothermic reactions at: brainly.com/question/13892884

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1 year ago

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2 years ago

How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __

kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

determine the number of moles of HCl
determine the mole ratio,
use the mole ratio to calculate the number of moles of aluminum.
use RFM of Aluminum to determine the grams required.

Moles of HCl

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

$x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \:$

We have 0.07 moles of HCl.

Mole ratio

Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

but moles of HCl is 0.07, therefore the moles of Al;

$= \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles$

Therefore we have 0.0233333 moles of aluminum.

Grams of Aluminum

We use the formula;

$grams \: = moles \: \times \: rfm$

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

$= 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams$

The number of grams of aluminum required to react with HCl is 0.6258 g.

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2 years ago