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3 years ago

6

Which reactant will be in excess, and how many moles of it will be left over?

2 answers:

olga_2 [115]3 years ago

7 0

Answer:

Oxygen and 0.36

Explanation:

Oxygen is usually always in excess and you use the math to find the 0.36

timofeeve [1]3 years ago

5 0

Answer:

Oxygen with 0.36 moles left over

Explanation:

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Given 6.0 mol of N2 are mixed with 12.0 mol of H2. Equation: N2 + 3H2--> 2NH3

andrey2020 [161]

1) Excess reagent

1 mol N2 / 3 mol H2

6.0 mol N2 *3 mol H2 / 1 mol N2 = 18 mol H2

18mol H2 > 12 mol H2 => H2 is limiting (you need 18 mol H2 to use all the 6 mol N2), then N2 is in excees.

12.0 mol H2 * 1mol N2/ 3 mol H2 = 4 mol N2 is the quantity that will react, then the excess is 6 mol N2 - 4 mol N2 = 2 mol N2

2) NH3 produced

12 mol H2 * [2 mol NH3 / 3 mol H2] = 8 mol NH3

Aslso, 4 mol N2 *[2molNH3 / 1 molN2] = 8 mol NH3, the same result.

3) Yield

80% * 8 mol NH3 = 6.4 mol NH3

5 0

3 years ago

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr

jenyasd209 [6]

Answer : The final temperature of the water is, $22.5166^oC$

Explanation : Given,

Mass of benzene = 8.600 g

Molar mass of benzene = 78 g/mole

First we have to calculate the moles of benzene.

$\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole$

Now we have to calculate the energy of combustion.

The given balanced chemical reaction is:

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ$

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives $\frac{6542 kJ}{2}\times 0.1103=360.7913kJ$ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : $q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})$

where,

$q_w$ = heat released = 360.7913 kJ = 36079.13 J

$m_w$ = mass of water = 5691 g

$c_w$ = specific heat of water= $4.18J/g^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $21^oC$

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

$36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)$

$T_{final}=22.5166^oC$

Therefore, the final temperature of the water is, $22.5166^oC$

8 0

4 years ago

Calculate how many grams of oxygen form when each quantity of reactant completely reacts.. . 2HgO----->2Hg+O2

goblinko [34]

<span>6.96g HgO
hope it helps.
</span>

3 0

3 years ago

When a student titrates a 5.00 mL of formic acid (HCHO2) with 26.59 mL of 0.1088 M NaOH, find the concentration of formic acid s

nikitadnepr [17]

Find it yourself my dude please

6 0

2 years ago

Find the pH of NaOH with a concentration of 0.0600 M.

Citrus2011 [14]

Answer:

12.78

Explanation:

pOH= -log(0.0600) = 1.22184875

pH= 14-1.22184875 = 12.78

5 0

3 years ago