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Nonamiya [84]
3 years ago
12

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
Chemistry
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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2 years ago
A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
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Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0
3 years ago
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When an aqueous solution containing hydrochloric acid is just neutralized with an aqueous solution containing sodium hydroxide,
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Answer:

c. salty.

Explanation:

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The reaction of an acid , i.e hydrochloric acid and a base i.e. , sodium hydroxide takes place , which is a type of neutralization reaction,

Where the acid and a base reacts to give salt along with water .

From the question,

The reaction is as follows -

2 HCl  +  NaOH  →  NaCl  +  H₂O  

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