If excess carbon disulfide reacts with 450 mL of oxygen, 150 mL of carbon dioxide and 300 mL of sulfur dioxide gases would be produced respectively.
<h3>Stoichiometric calculation</h3>
The reaction between liquid carbon disulfide and oxygen is represented by the equations below:
The mole ratio of oxygen to carbon dioxide and sulfur dioxide produced is 3:1:2.
Thus, for 450 mL oxygen, 1/3 x 450 = 150 mL of carbon dioxide will be required.
Also for 450 mL of oxygen, 2/3 x 450 = 300 mL of sulfur dioxide will be required.
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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To determine the mass, you need to know the molecular weight of the c8h10n4o2 . The molecular weight of <span>c8h10n4o2 would be: 8*12 + 10*1 + 4*14 + 2*16= 194g/mol.
To convert the number of molecules into moles, you need to divide it with 6.02 * 10^23. The calculation of the mass of </span>c8h10n4o2 would be:
(7.20×10^20 molecules) /(6.02 * 10^23 molecule/mol) * 194g/mol= 232 * 10^-3 grams= 0.232 grams
Answer:
Explanation:
We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.
We will convert using dimensional analysis and set up a ratio using Avogadro's Number.
We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.
Flip the ratio so the units of atoms of fluorine cancel each other out.
Condense into 1 fraction.
Divide.
The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.
4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
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To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
