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aleksklad [387]
3 years ago
5

If 42 grams of carbon and 52 grams of oxygen are used, how many grams of CO2 will be produced (Hint: find the

Chemistry
1 answer:
zavuch27 [327]3 years ago
8 0

Answer:

71.5g

Explanation:

The reaction equation is given as:

               C  +  O₂  →  CO₂

Mass of C = 42g

Mass of O₂  = 52g

Unknown:

Mass of CO₂ produced  = ?

Solution

Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.

 The extent of the reaction is controlled by this reactant.

Find the number of moles of the given species;

 Number of moles  = \frac{mass}{molar mass}

      Number of moles of C  = \frac{42}{12}   = 3.5mol

     Number of moles of O₂   = \frac{52}{32}   = 1.63mol

Now;

   From the balanced reaction equation;

           1 mole of C reacted with 1 mole of O₂

We see that C is in excess and O₂ is the limiting reactant.

            1 mole of O₂ will produce 1 mole of CO₂

 So;      1.63mole of O₂ will produce 1.63 mole of CO₂

Mass of CO₂ = number of moles x molar mass

       Molar mass of CO₂ = 44g/mol

Mass of CO₂ = 1.63 x 44 = 71.5g

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a) 2Na(s) +2H_2O(I)→ 2Na^+(aq) + OH^-(aq)+ H_2(g)

b) 2Ag (s) +2H^(aq) → 2 Ag^+ (aq) +H_2(g)

<h3>What are half-reactions?</h3>

The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.

a)

2Na(s) +2H_2O(I)→ 2Na^+(aq) + OH^-(aq)+ H_2(g)

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2 years ago
) B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 react
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<u>Answer:</u> The amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

<u>Explanation:</u>

The chemical equation for the reaction of B_5H_9 with oxygen gas follows:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ

To calculate the amount of heat released for the given amount of B_5H_9(l), we use unitary method, we get:

When 2 moles of B_5H_9(l) reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of B_5H_9(l) will react, the amount of heat released will be = \frac{5259}{2}\times 0.211=554.8kJ

Hence, the amount of heat released when 0.211 moles of B_5H_9(l) reacts is 554.8 kJ

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