Table salt, NaCl(s), and also sugar, C12H22O11(s), are accidentally mixed. A 4.50-g sample is burned, and also 2.90 g the CO2(g) is produced. What is the mass portion of the table salt in the mixture?


C12H22O11 + O2 → CO2 + H2O ...

Now balance it:


C12H22O11 + 12O2 → 12CO2 + 11H2O

molar mass CO2 = 12.01 + 16.00 x 2 = 44.01 g/mol

moles CO2 developed = 2.90 g CO2 x <1 mol / 44.01 g> = 0.0659 mol CO2

From the balanced equation the mole ratio C12H22O11 : CO2 = 1 : 12

so assuming all the C12H22O11 in the sample is burned 0.0659/12 = 0.00549 mol C12H22O11 room in the sample

molar fixed C12H22O11 = 12.01 x 12 + 1.01 x 22 + 16.00 x 11 = 342.34 g/mol

mass C12H22O11 in the sample = 0.00549 mol C12H22O11 x <342.34 g/mol> = 1.88 g C12H22O11

so massive NaCl in the sample = 4.50 - 1.88 = 2.62 g NaCl

mass percent NaCl = <2.62 g NaCl / 4.50 g> x 100% = 58.2% NaCl


Table Salt Nacl


I"m not sure around this and also hope someone rather can examine it out for you.

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Salt will not burn under plain conditions, so when the sample to be "burned", only the sugar to be burned, the salt to be left behind.

So: C12H22O11 + 35/2 O2 ------------------> 12 CO2 + 11 H2O

Now 2.90 mol the CO2 is same to 2.90 / 44.01 g/mol = 0.06589 mol


According to the equation the quantity of sugar the will develop that variety of moles is 1/12 of the number of moles that CO2.

So street is 1/12 x 0.06589 = 0.0054911 mol of sugar.

The massive of that sugar is 0.0054911 mol x 342.34 g/mol = 1.88 g

So out of the sample that the mixture 1.88 g of it was sugar. That leaves 4.50 - 1.88 = 2.62 g of salt

So the mass percent that salt in the mixture would be 2.62 / 4.50 all x 100 = 58.2%


Jamian"s answer is correct because that the most part, however to complete the problem and find the fixed percent, you actually subtract the mass of NaCl (2.14g) indigenous the original sample, 5.50g. Mass of NaCl in mixture: 5.50g - 2.14g = 3.36g NaCl in sample To find mass percent, just divide (3.36g NaCl)/(5.50g full mass the sample) = 0.6109 so the fixed % of NaCl in the mixture is 61.09%.

See more: Which One Is Correct, 'Do You Meant' Or ' Did You Mean: Grammar


Idk I"m in 8 hours grade hahaha


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