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4 years ago

When 7.0 mol Al react with 8.5 mol HCl, what is the limiting reactant and how many moles of AlCl3 can be formed?

2 answers:

4 0

Option 2 is Correct.
because, in the standard reaction, 2 moles of Al reacts with 6 of moles of HCl.
if 7 moles of Al are there, it will require 21 mole of HCl. So HCl Is the limiting reagent.

Now, 6 moles of HCl gives 2 moles of AlCl3, then 8.5 moles of HCl will give 2.84 moles of AlCl3.

crimeas [40]4 years ago

4 0

B. HCl is the limiting reactant; 2.8 mol AlCl3 can be formed

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Describe the trends in electron configuration in the periodic table by selecting the terms from the drop-down menus.

natka813 [3]

1)Left to right in a period it goes up by one
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5 0

3 years ago

Read 2 more answers

Determine the moles of chlorine needed to react with 38.9 of iron (III) chloride.

lys-0071 [83]

Answer: 0.360 moles

Explanation:

To calculate the moles :

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

$\text{Moles of} FeCl_3=\frac{38.9g}{162.2g/mol}=0.240moles$

The balanced chemical reaction is:

$2Fe+3Cl_2\rightarrow 2FeCl_3$

According to stoichiometry :

2 moles of require = 3 moles of

Thus 0.240 moles of $FeCl_3$ will require= $\frac{3}{2}\times 0.240=0.360moles$ of $Cl_2$

Thus 0.360 moles of chlorine needed to react to produce 38.9 of iron (III) chloride.

6 0

3 years ago

Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is

mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

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

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

$\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol$

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat = $\frac{-9078.57}{2}=-4539.28kJ$

For per gram of compound:

Molar mass of $B_5H_9$ = 63.12 g/mole

$\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g$

Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

7 0

3 years ago

A.1 B.2 C.3 D.4 Please help ASAP

maks197457 [2]

Answer:

Explanation:

it states 3 elements names so im assuming thats it and not including the element symbols

8 0

3 years ago

If 585.24 Joules of heat are added to 53.2 grams of water at 24.15oC, what will the new temp. be?

spayn [35]

E = mct
Energy = (mass) x (specific heat capacity of water) x (change in temp)

585.24 = 53.2 x 4.2 x (X-24.15)

585.24 divided by 53.2 divided by 4.2 = X - 24.15

2.62 = X - 24.15

X= 26.77degrees C

(Specific heat capacity for water is 4.2 but is different for other liquids)

3 0

4 years ago