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Lostsunrise [7]
2 years ago
6

The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, in grams, must react to produce 50.

00 g of O2?
Chemistry
1 answer:
snow_lady [41]2 years ago
5 0

112.5 g. The production of 50.00 g O2 requires 112.5 g H2O.

a) Write the partially balanced equation for the decomposition of water.

MM = 18.02 32.00

2H2O → O2 + …

Mass/g = 50.00


b) Calculate the <em>moles of O2
</em>

Moles of O2 = 50.00 g O2 × (1 mol O2/16.00 g O2) = 3.1250 mol O2


c) Calculate the <em>moles of water</em>

Moles of H2O = 3.1250 mol O2 × (2 mol H2O/1 mol O2)

= 6.2500 mol H2O


d) Calculate the mass of water

Mass of H2O = 6.2500 mol H2O × (18.02 g H2O/1 mol H2O)

= 112.5 g H2O

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c

for more information please visite:

brainly.com/question/15017356?referrer=search

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2 years ago
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Mashutka [201]
What is it asking for?
6 0
2 years ago
Read 2 more answers
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

7 0
3 years ago
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