The balanced equation for the above reaction is as follows:
2H2 + O2 = 2H2O
From the above equation it is clear that 2 mole H2 react with 1 mole O2
Molar mass of H2 = 2g
Molar mass of O2= 32 g
This implies,
4 g H2 react with 32 g O2
3 g H2 reacts with = (32/4) x 3g of O2 gas= 24 g
As the given amount of O2 is more than required therefore O2 is excess reagent and H2 is limiting reagent.
2 mole of hydrogen gas reacts to form 2 mole of water molecule therefore,4 g of H2 produces = 36 g of water So the amount of H2O produced by 3 g H2 = (36/4) x 3= 27 g
Hence, 27 g of water will be produced during the recation
As, 24 g of oxygen has been utilised during the recation and 29 g of oxygen was supplied therefore amount of oxygen gas left is (29-24) = 5g
Abby A.
1 Expert Answer
It seems as though I am doing all of your homework for you. So, in this problem, I'll get you started and hopefully you can complete the assignment.
Always start with the balanced equation:
2H2 + O2 ==> 2H2O ... balanced equation
Anytime you are given amounts of BOTH reactants, you MUST find the limiting reactant:
10.9 g H2 x 1 mol H2/2 g H2 = 5.45 mol H2
16.03 g O2 x 1 mol O2/32 g O2 = 0.500 mol O2
Clearly, by looking at the balanced equation you can tell that O2 is limiting because it takes 2 mol H2 for each 1 mol O2, and there is plenty of H2 present (5.45 mol vs. 0.500 mol).
Now you can find maximum amount H2O formed:
0.500 mol O2 x 2 mol H2O / 1 mol O2 x 18 g H2O/mol H2O = answer in grams
Answer
Text Solution
Solution : The balanced chemical equation is : <br> `underset(2xx2=4g)(2H_(2))+underset(32g)(O_(2))rarrunderset(2xx18=36 g)(2H_(2)O)` <br> Step I. Determination of limiting reactant <br> 4 g of `H_(2)` react with `O_(2)=32` g <br> `:.` 3 g of `H_(2)` react with `O_(2)=(32)/(4)xx3g = 24 g` <br> But the amount of oxygen actually available = 29 g<br> As oxygen is available in excess therefore, hydrogen is the limiting reactant. <br> Step II. Calculation of the amount of water that can be formed <br> 4 g of `H_(2)` in the reaction = 36 g <br> `:.` 3 g of `H_(2)` require `O_(2)` for reaction `= (32)/(4) xx 3g = 24 g` <br> But oxygen which is actually present = 29 g <br> `:.` Amount of oxygen left unreacted `= 29 - 24 = 5g`.
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3 g of H 2 react with 29 g of O 2 to yield H 2 O . which is the limiting reactant? Calculate the maximum amount of H 2 O that can be formed. Calculate the amount the reactant which remains unreacted.
As the given amount of O2 is more than required therefore O2 is the excess reagent and H2 is the limiting reagent. 2 mole of hydrogen gas reacts to form 2 mole of the water molecule, therefore,
4 g of H2 produces = 36 g of water
So the amount of H2O produced by 3 g H2 = (36/4) x 3
= 27 g Hence, 27 g of water will be produced during the reactionAs 24 g of oxygen has been utilized during the reaction and 29 g of oxygen was supplied therefore the amount of oxygen gas left is (29-24) = 5g
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