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klasskru [66]
3 years ago
9

Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much sulfur dioxide

would you need to completely react with 6.00 g O2 such that all reactants could be consumed?
Chemistry
1 answer:
taurus [48]3 years ago
7 0
1) Balanced chemical equation:

2SO2 (g) +  O2 (g) -> 2SO3 (l)

2) Molar ratios

2 mol SO2 : 1 mol O2 : 2 mol SO3

3) Convert 6.00 g O2 to moles

number of moles = mass in grams / molar mass

number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.

4) Use proportions with the molar ratios

=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2

=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.

5) Convert 0.375 mol SO2 to grams

mass in grams = number of moles * molar mass

molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol

=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g

Answer: 24.0 g of SO2 are needed to react completely with 6.00 g O2.
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Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

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Answer:

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How many moles of atoms are in 376.2 g S?
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