Question

1. Sodium and water react according to the following equation. If 31.5g of sodium are added to excess water, how many liters of hydrogen gas are formed at STP?

2Na+2H2O—>2NaOH+H2

2. Sodium reacts with oxygen to produce sodium oxide as described by the balanced equation below. If 54.1g of sodium reacts with excess oxygen gas to produce 61.8g Of sodium oxide, what is the percent yield? (Hint: be sure to calculate theoretical yield first)
4Na+O2–>2Na2O

(Please show work)

Answer 1

Question 1

The number of liter of hydrogen gas that are formed at STP is

15.344 L

calculation

2 Na +2 H₂O → 2NaOH +H₂

calculate the moles of Na

moles =mass/molar mass

from periodic table the molar mass of Na = 23 g/mol

moles = 31.5 g /23 g /mol = 1.37 moles

From equation above the mole ratio of Na: H2 is 2:1

therefore the moles of H2 =1.37 x1/2 =0.685 moles

At STP 1 moles of a gas = 22.4 L

0.685 moles = ? L

by cross multiplication

=[ (0.685 moles x 22.4 L) / 1 mole =15.344 L

Question 2

The percent yield = 84.77%

calculation

4 Na + O₂ → 2 Na₂O

%yield =actual yield / theoretical yield x 100

Actual yield= 61.8 g

The theoretical yield is calculated as below

Step 1: find the moles of Na

moles = mass /molar mass

from periodic table the molar mass of Na = 23 g/mol

moles = 54.1 g /23 g/mol =2.352 moles

Step 2: use the mole ratio to determine the moles of Na₂O

Na:Na₂O is 4: 2 therefore the moles of Na₂O = 2.352 x 2/4 =1.176 moles

Step 3: find the theoretical mass of Na₂O

mass = moles x molar mass

The molar mass of Na₂O = [(23 x 2 ) + 16] =62 g/mol

mass = 1.176 moles x 62 g/mol = 72.9 g

% yield is therefore = 61.8 g /72.9 x 100 = 84.77%