Practice Problem 9.85 Which reactant is limiting? Express your answer as a chemi

Question

Practice Problem 9.85 Which reactant is limiting? Express your answer as a chemical formula. Sodium (Na) reacts with hydrogen (H2) to form sodium hydride (NaH). A reaction mixture contains 10.84 g Na and 2.25x102 g H2 Submit Mv Answers Give U Correct Part C What is the theoretical yield for this reaction in grams? Express your answer with the appropriate units. mNaH 0.480 Submit My Answers Give Up Incorrect; Try Again Part D How many grams of excess reactant are left over at the end of the reaction? Express your answer with the appropriate units. mNaValue Units Submit My Answers Give Up Part E When this reaction is actually performed, 0.428 g of NaH is recovered. What is the percent yield of the reaction? Activate Windows Go to Settings to activate Windows. Submit My Answers Give Up

Explanation / Answer

PART B:

Sodium (Na) reacts with hydrogen (H2) to form sodium hydride (NaH). It is represented as

Na     +     H2     ------->     NaH

The balanced equation is represented as
2Na     +     H2     ------->     2NaH

In the above balanced reaction equation;

2 moles of Na reacts with 1 mole of H2.

Now,

10.84 g of Na
Molar mass of Na = 23 g/mol
So, 23 g of Na = 1 mol
1 g of Na = (1/23) mol
10.84 g of Na = (10.84 / 23) mol = 0.47 mol

2.25 x 10-2 g of H2
Molar mass of H2 = 2 g/mol
So, 2 g of H2 = 1 mol
1 g of H2 = (1/2) mol
(2.25 x 10-2) g of Na = (2.25 x 10-2)/2 mol
                                  = 1.125 x 10-2 mol
                                  = 0.01125 mol

Since, moles of H2 is less, so H2 is the limiting reagent.

PART C:

2Na     +     H2     ------->     2NaH

Since, H2 is the limiting reagent and in the above balanced reaction equation;
1 mole of H2 produces 2 moles of NaH
0.01125 moles of H2 produce (0.01125 x 2) moles of NaH
0.01125 moles of H2 produce 0.0225 moles of NaH.

Molar mass of NaH = 24 g/mol
So, 1 mole of NaH = 24 g
0.0225 mole of NaH = 0.0225 x 24 g = 0.54 g (theoretical yield)

PART D:

2Na     +     H2     ------->     2NaH

In the above balanced reaction equation;
2 moles of Na reacts with 1 mole of H2.

Moles of Na = (10.84 / 23) mol = 0.47 mol
Moles of H2 = 0.01125 mol

Since, H2 is limiting reagent. So, moles of Na reacted = 2 x 0.01125 moles = 0.0225 moles
Moles of unreacted Na = 0.47 mol - 0.0225 mol = 0.4475 mol

Molar mass of Na = 23 g/mol
So, 1 mol of Na = 23 g
0.4475 mol of Na = 0.4475 x 23 mol = 10.29 g

PART E:

Practical or actual yield = 0.428 g
Theoretical Yield = 0.54 g
% Yield = (Practical yield / Theoretical Yield) x 100
              = (0.428 g / 0.54 g) x 100
              = 79.26 %

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