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Consider the equation. 2H2(g) + O2(g) >>>> 2H2O(g) Which represents the correct mass-volume relationship at STP?
A.) 32.00 g of O2 react with an excess of H2 to produce (2 ´ 22.4) L of H2O.
B.) 2.00 g of H2 react with an excess of O2 to produce (2 ´ 22.4) L of H2O.
C.) 32.00 g of O2 react with an excess of H2 to produce 22.4 L of H2O.
D.) (2 ´ 2.00) g of H2 react with an excess of O2 to produce 22.4 L of H2O.

Respuesta :

Edufirst
The equation  2H2(g) + O2(g) ---> 2H2O(g) , gives you the theoretical molar ratios of reactants and products:


2 mol H2(g) : 1 mol O2(g) : 2 mol H2O (g)


You know that at SPT 1 mol of gas is equivalent to 22.4 liter of gas, so you also can write these volumetric ratios:  


2*22.4 liter H2(g) : 22.4 litr O2(g) : 2*22.4 liter H2O (g)



Also , using the molar masses you can state the mass ratios:


2*2* 1 g H2: 2*16 g O2 : 2*18 g H2O

equivalent to 4 g H2 : 32 g O2 : 36 g H2O

When you use the information from the three set of ratios you get to

32.0 g O2 (equivalent to 1 mol and 22.4 liter) react with excess H2 to produce 36 g H2O (equivalent to 2 moles and 2*22.4 liter), which is the option A.


Answer: option A. 32 g O2 react with an excess H2 to produce 2*22.4 liter H2O  

Answer:

it’s A

Explanation:

taking it rn

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