Use 1 decimal point for all atomic masses. 2.00 g of H2(g) are reacted with 32.0 g of O2(g) by the following reaction 2H2(g) + O2(g) --> 2H2O(g)What is the limiting reagent?H2(g)O2(g)    Based on the limiting reagent, fill in the following ICE Table using moles. Enter any zero values as 0.00.  H2O2H2OInitial   Change   Final    How much of the excess reagent remains (in grams)?

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Asked Nov 13, 2019
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Use 1 decimal point for all atomic masses. 2.00 g of H2(g) are reacted with 32.0 g of O2(g) by the following reaction

 
2H2(g) + O2(g) --> 2H2O(g)
What is the limiting reagent?
H2(g)
O2(g)    
Based on the limiting reagent, fill in the following ICE Table using moles. Enter any zero values as 0.00.

 

  H2 O2 H2O
Initial      
Change      
Final      

 

How much of the excess reagent remains (in grams)?

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Expert Answer

Step 1

The balanced reaction is written below.

2H2 (g) + O2 (g) → 2H2O (g)

The number of moles of H2 is calculated in equation (1).

The number of moles of O2 is calculated in equation (2).

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n М 2.00 g .... (1) 2.0 g/mol 1mol m n М 32.0g 32.0 g/mol (2) 1 mol

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Step 2

According to the balanced chemical reaction, 2 mol of H2 requires 1 mol of O2. Thus, the moles of oxygen required by 1 mol of H2 is calculated below.

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1mol x1 mol 2 mol = 0.5 mol

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Step 3

The number of moles of oxygen actually present is 1 mol. Thus, O2 is pre...

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