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3 years ago

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The molar mass of H2O is 18.01 g/mol. The molar mass of O2is 32.00 g/mol. What mass of H2O,in grams, must react to produce 50.00

g of O2

2 answers:

N76 [4]3 years ago

7 0

Answer is: mass of water is 56.28 grams.
Chemical reaction: 2H₂O → 2H₂ + O₂.
m(O₂) = 50.00 g.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 50 g ÷ 32 g/mol.
n(O₂) = 1.5625 mol.
From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.
n(H₂O) = 2 · 1.5625 mol.
n(H₂O) = 3.125 mol.
m(H₂O) = n(H₂O) · M(H₂O).
m(H₂O) = 3.125 mol · 18.01 g/mol.
m(H₂O) = 56.28 g.

Svet_ta [14]3 years ago

3 0

Answer:

56.28 grams.

Explanation:

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Answer;

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

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Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

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Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

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Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

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Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

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Explanation:

a. A liquid that dissolves another substance is called the solvent.

b. A chemical that is dissolved solute.

c. A value used to describe the amount of one substance dissolved in another is called concentration.

d. A liquid consisting of one substance dissolved in another is called a solution.

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