(I corrected it )
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
My work: For #1 Na would have 2 moles and H would have 1? Make 31.5g x 2/1=63? Or 31.5g divided by 2 x 1= 15.7?
(If im wrong can you help me)
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
I’m having trouble solving #2 can someone give me a format on how to solve for the theoretical yield.
I know how to solve the problem just not the theoretical yield that’s the part I’m struggling on.
I’m not sure but I think the equation is already balanced. (If not can someone tell me so I can correctly try to fix it)
2 answers
moles of Na = 31.5 / 23.0
moles of H2 is half of Na
... from reaction equation
a mole of ANY gas is 22.4 L at STP
2. the equation is balanced
theoretical is what you get if ALL the sodium reacts with oxygen to form sodium oxide
54.1/23.0 moles of Na, theoretically yields 54.1/46.0 moles of Na2O
... molar mass of Na2O is 62.0 g
find the theoretical product mass and compare it to the actual to get the % yield