sodium metal reacts with hydrogen gas to yield solid sodium hydride ( NaH ). A reaction mixture contains 10.00 g sodium metal and 0.235 g of hydrogen gas.
write a balanced chemical equation for this reaction?
what is your limiting reactant?
what is your theoretical yield (in grams)?
how many grams of excess reactant are left over at the end of the reaction?
if 4.28 g of NaH were recovered at the end of the reaction, what is the percent yield of the reaction?
Note: When you show NO EFFORT I usually don't work these. Surely you know SOMETHING about this. Surely the equation.
sodium metal reacts with hydrogen gas to yield solid sodium hydride ( NaH ). A reaction mixture contains 10.00 g sodium metal and 0.235 g of hydrogen gas.
write a balanced chemical equation for this reaction?
2Na + H2 ==> 2NaH
what is your limiting reactant?
mols Na = grams/atomic mass Na = ?
mols H2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Na to mols NaH product.
Do the same and convert mols H2 to mols NaH product.
The limiting reagent (LR) is the reagent (Na or H2) that produces the smaller number of mols of the product.
what is your theoretical yield (in grams)?
Take the smaller number of mols x molar mass NaH to find the grams of NaH produced. This is the theoretical yield (TY).
how many grams of excess reactant are left over at the end of the reaction?
Using the coefficients in the balanced equation, convert mols LR to mols excess reagent (ER) USED, that times atomic or molar mass to get grams. Then subtract from the grams initially to find how much is left over.
if 4.28 g of NaH were recovered at the end of the reaction, what is the percent yield of the reaction?
The actual yield (AY) is 4.28 grams.
%yield = (AY/TY)*100 = ?
Post your work if you have trouble.
PLEASE HELP
sodium metal reacts with hydrogen gas to yield solid sodium hydride ( NaH ). A reaction mixture contains 10.00 g sodium metal and 0.235 g of hydrogen gas.
2Na + H2 ==> 2NaH
what is the limiting reactant?
what is the theoretical yield (in grams)?
I gave you a DETAILED procedure for working this problem in toto. Step by step, explain, in detail, what you don't understand. As far as I can tell you've just rewritten the problem.
To write the balanced chemical equation for the reaction, we need to ensure that the number of atoms on both sides of the equation is equal.
The balanced equation for the reaction between sodium metal (Na) and hydrogen gas (H₂) to form sodium hydride (NaH) is:
2Na + H₂ → 2NaH
To determine the limiting reactant, we can compare the moles of each reactant to the stoichiometry of the balanced equation.
First, let's calculate the moles of each reactant:
Moles of sodium (Na) = mass / molar mass = 10.00 g / 22.99 g/mol = 0.435 mol
Moles of hydrogen (H₂) = mass / molar mass = 0.235 g / 2.02 g/mol = 0.116 mol
Next, compare the moles of each reactant to the stoichiometry of the balanced equation. From the equation, we can see that the stoichiometric ratio between Na and H₂ is 2:1. Therefore, we need twice as many moles of Na as H₂ to fully react.
In this case, the moles of Na (0.435 mol) are more than twice the moles of H₂ (0.116 mol). This means that the hydrogen gas (H₂) is the limiting reactant.
To calculate the theoretical yield, we need to determine the number of moles of the limiting reactant (H₂) and convert that to grams of NaH using the stoichiometry of the balanced equation.
From the balanced equation, 1 mole of H₂ reacts to produce 2 moles of NaH.
Moles of NaH = (0.116 mol H₂) × (2 mol NaH / 1 mol H₂) = 0.232 mol NaH
The molar mass of NaH is 22.99 g/mol + 1.01 g/mol = 23.99 g/mol.
The theoretical yield (in grams) of NaH is:
Theoretical yield = Moles of NaH × Molar mass of NaH
= 0.232 mol × 23.99 g/mol
= 5.69 g
To determine the grams of excess reactant left over at the end of the reaction, first, we need to calculate the moles of the excess reactant (Na) that did not react.
Moles of Na left over = Moles of Na - (Moles of NaH formed / 2)
= 0.435 mol - (0.232 mol / 2)
= 0.319 mol
The mass of the excess reactant left over can now be calculated:
Mass of Na left over = Moles of Na left over × Molar mass of Na
= 0.319 mol × 22.99 g/mol
= 7.33 g
To calculate the percent yield of the reaction, we compare the actual yield (4.28 g NaH) to the theoretical yield (5.69 g NaH).
Percent yield = (Actual yield / Theoretical yield) × 100
= (4.28 g / 5.69 g) × 100
= 75.2% (rounded to one decimal place)
Therefore, the percent yield of the reaction is 75.2%.