When 3g of H2 reacts with 29g of O2 the amount of h2o formed is?

For starters, you need a balanced chemical equation that describes this reaction

So, you know that every mole of oxygen gas that takes part in the reaction consumes #2# moles of hydrogen gas and produces #2# moles of water.

The problem provides you with the masses of the two reactants, so use their respective molar mass to convert this to moles

#3 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "1.488 moles H"_2#

#29 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.906 moles O"_2#

Now, your goal here is to figure out if you have enough moles of hydrogen gas present to ensure that all the moles of oxygen gas take part in the reaction.

In your case, you can say that #0.906# moles of oxygen gas would require

#0.906 color(red)(cancel(color(black)("moles O"_2))) * "2 moles H"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = "1.812 moles H"_2#

SInce you have fewer moles of hydrogen gas than you would need

#overbrace("1.488 moles H"_2)^(color(blue)("what you have")) " " < " " overbrace("1.812 moles H"_2)^(color(blue)("what you need"))#

you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of oxygen will get the chance to react.

This means that the reaction will consume #1.488# moles of hydrogen gas and

#1.488 color(red)(cancel(color(black)("moles H"_2))) * "1 mole O"_2/(2color(red)(cancel(color(black)("moles H"_2)))) = "0.744 moles O"_2#

and produce

#1.488 color(red)(cancel(color(black)("moles H"_2))) * ("2 moles H"_2"O")/(2color(red)(cancel(color(black)("moles H"_2)))) = "1.488 moles H"_2"O"#

To convert this to grams, use the molar mass of water

#1.488 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("27 g")))#

I'll leave the answer rounded to two sig figs, but do not forget that you only have one significant figure for the mass of hydrogen gas.

After the reaction is complete, you will be left with

#"0.906 moles O"_2 - "0.744 moles O"_2 = "0.162 moles O"_2#

To convert this to grams, use the molar mass of oxygen gas

#0.162 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("5.2 g")))#

Once again, I'll leave the answer rounded to two sig figs.

Further,

4 g H2 reacts with 32 g O2

3 g H2 reacts with (32/4) x 3 g of O2 gas

= 24g

H2 is the limiting reagent and O2 is the excess reagent since the amount of O2 provided is greater than what is needed. 2 moles of hydrogen gas combine to generate 2 moles of water as a result.

4 g of H2 produces 36 g of water

Amount of H2O produced by 3 g H2 = (36/4) x 3 = 27g

Therefore, the reaction will result in the production of 27 g of water. Since 29 g of oxygen was provided and 24 g was used in the reaction, the remaining amount of oxygen gas is (29-24) = 5g

Summary:

3g of H2 react with 29g of O2 to yield H2O the limiting reactant is hydrogen. The maximum amount of water that can be formed is 27 g. The amount of the reactant which remains unreacted is 5g.

Text Solution

Solution : The balanced chemical equation is : `underset(2xx2=4g)(2H_(2))+underset(32g)(O_(2))rarrunderset(2xx18=36 g)(2H_(2)O)` Step I. Determination of limiting reactant 4 g of `H_(2)` react with `O_(2)=32` g `:.` 3 g of `H_(2)` react with `O_(2)=(32)/(4)xx3g = 24 g` But the amount of oxygen actually available = 29 g As oxygen is available in excess therefore, hydrogen is the limiting reactant. Step II. Calculation of the amount of water that can be formed 4 g of `H_(2)` in the reaction = 36 g `:.` 3 g of `H_(2)` require `O_(2)` for reaction `= (32)/(4) xx 3g = 24 g` But oxygen which is actually present = 29 g `:.` Amount of oxygen left unreacted `= 29 - 24 = 5g`.

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