How many grams of excess reactant are left over at the end of the reaction? When

Question

How many grams of excess reactant are left over at the end of the reaction? When this reaction is actually performed, 12.50 g of ClF_3 is recovered. What the percent yield of the reaction? Sodium (Na) reacts with hydrogen (H_2) to form sodium hydride (NaH). A reaction mixture contains 10.00 g Na and 0.0235 g H_2. Write a balanced chemical equation for this reaction. Which reactant is limiting? What is the theoretical yield for this reaction in grams? How many grams of excess reactant are left over at the end of the reaction? When this reaction is actually performed, 0.428 g of NaH is recovered. What is the percent yield of the reaction? Butane (C_4 H_10) used as the fuel in disposable lighters, reacts with oxygen (O_2) to produce CO_2 and H_2 O. Suppose 10.00 g butane is combined with 10 00 g O_2. Write a balanced chemical equation for the combustion of butane.

Explanation / Answer

m = 10 g of Na and 0.0235 g of H2

a) balanced equation

2Na + H2 --> 2NaH

b)

find lmiting reactant

MW of NA = 23

MW of H2 = 2

change to mol

mol Na = mass/MW = 10/23 = 0.4347 mol of Na

mol H2 = mass/MW = 0.0235 /2 = 0.0118 mol of H2

H2 is clearly the limitin greactant

c)

%yield

if H2 is the limiting.. then

1 mol of H2 --> 2 mol of NaH

0.0118 --> 2*0.0118 = 0.0236 mol of NaH

MW of NaH = 24

mass = mol*MW = 24*0.0236 = 0.5664 grams of NaH

d)

excess of Na?

inital mol of Na = 0.4347 mol

reaction mol of Na = 2*0.0118 = 0.0236 mol

final mol Na = 0.4347 -0.0236 = 0.4111 mol of NA

mass = mol*MW = 0.4111 *23 = 9.4553 grams of Na are left

e)

m = 0.428 g

% yield = real/theretical * 100 = 0.428 /0.5664 *100 = 75.56%

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