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kondaur [170]
4 years ago
11

For the reaction ? C6H6 + ? O2 → ? CO2 + ? H2O 42.5 grams of C6H6 are allowed to react with 113.1 grams of O2. How much CO2 will

be produced by this reaction? Answer in units of grams
Chemistry
1 answer:
ziro4ka [17]4 years ago
3 0

Answer:

There will be 143,67g CO2 produced

Explanation:

2 C6H6 + 15 O2 → 12 CO2 + 6 H2O

(42,5 g C6H6) / (78.1124 g C6H6/mol) = 0.54408775 mole C6H6

(113.1 g O2) / (31.9989 g O2/mol) = 3.534496 moles O2

0.54408775 mole of C6H6 would react completely with 0.54408775 x (15/2) = 4.080658 mole O2, but there is more O2 present than that, so O2 is in excess and C6H6 is the limiting reactant.

(0.54408775 mol C6H6) x (12/2) x (44.0096 g/mol) = 143.67 g CO2

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Calculate the molarity of 13.1 g of NaCl in 727 mL of solution
valentina_108 [34]

Answer:

The molarity of the solution is 0,31 M

Explanation:

We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)

Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g

58, 5 g-----1 mol NaCl

13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl

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x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>

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4 years ago
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3 years ago
Consider 2H2 + O2 → 2H2O. To produce 1.2 g water, how many grams of H2 are required? Report to the correct number of significant
Elden [556K]

Answer:

0.133 mol (corrected to 3 sig.fig)

Explanation:

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no. of moles = mass / molar mass

so no. of moles of H2O produced = 1.2 / (1.0x2+16.0)

= 0.0666666 mol

From the equation, the mole ratio of H2:H2O = 2:2 = 1:1,

meaning every 1 mole of H2 reacted gives out 1 mole of water.

So, the no, of moles of H2 required should equal to the no, of moles of H2O produced, which is also  0.0666666 moles.

mass = no. of moles x molar mass

hence,

mass of H2 required = 0.066666666 x (1.0x2)

= 0.133 mol (corrected to 3 sig.fig)

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