SrH2 (s)+ 2H2O (l) >>> Sr(OH)2 (s) + 2H2 (g)
Calculate the mass of hydrogen gas that can be prepared from 5.22 grams SrH2 and 4.35 grams of H2O.
I keep getting 0.729 grams but its wrong...
SrH2 + 2H2O ==> Sr(OH)2 + 2H2
I suspect this is a limiting reagent type problem.
Convert grams to moles.
moles SrH2 = g/molar mass = 5.22/89.64 = 0.0582
moles H2O = 4.35/18.015 = 0.241
Convert moles of each to moles of product.
moles H2 from SrH2 = 0.0582 x (2 moles H/1 mole SrH2) = 0.0582 x (2/1) = 0.116.
moles H2 from H2O = 0.241 x (2 moles H2/2 moles H2O) = 0.241 x (2/2) = 0.241 x 1 0.241
Both answer, of course, can't be correct (because they are different); the correct one is ALWAYS the smaller. In this case, moles H2 produced is 0.116 and SrH2 is the limiting reagent.
Then convert 0.116 moles H2 to grams.
g = moles x molar mass = 0.116 x 2.016 = 0.23386 which rounds to 0.234 to three significant figures.
I tried two or three ways to get your answer of 0.769 and couldn't so I don't know where you went wrong.
Check my arithmetic.
To calculate the mass of hydrogen gas that can be produced from the given reactants, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
First, we need to calculate the number of moles for each reactant:
1. Calculate the moles of SrH2:
Molar mass of SrH2 (Sr = 87.62 g/mol, H = 1.01 g/mol)
Moles of SrH2 = mass / molar mass
Moles of SrH2 = 5.22 g / (87.62 g/mol + 2*1.01 g/mol)
Moles of SrH2 = 0.0561 mol
2. Calculate the moles of H2O:
Molar mass of H2O (H = 1.01 g/mol, O = 16.00 g/mol)
Moles of H2O = mass / molar mass
Moles of H2O = 4.35 g / (2*1.01 g/mol + 16.00 g/mol)
Moles of H2O = 0.232 mol
Next, we need to compare the moles of SrH2 and H2O to determine the limiting reactant. The balanced equation tells us that the ratio of SrH2 to H2O is 1:2.
3. Divide the number of moles of SrH2 by its stoichiometric coefficient:
Moles of SrH2 in excess = 0.0561 mol / 1 = 0.0561 mol
4. Divide the number of moles of H2O by its stoichiometric coefficient:
Moles of H2O in excess = 0.232 mol / 2 = 0.116 mol
Based on the stoichiometry, we can see that the moles of H2O are less than half of the moles of SrH2, indicating that H2O is the limiting reactant.
5. Calculate the moles of hydrogen gas produced using the moles of H2O:
Moles of H2 produced = Moles of H2O in excess * 2 (from the balanced equation)
Moles of H2 produced = 0.116 mol * 2 = 0.232 mol
Finally, calculate the mass of hydrogen gas using the moles of H2 and its molar mass:
6. Calculate the mass of H2:
Molar mass of H2 = 2*1.01 g/mol = 2.02 g/mol
Mass of H2 = Moles of H2 * molar mass
Mass of H2 = 0.232 mol * 2.02 g/mol
Mass of H2 = 0.469 g
The correct mass of hydrogen gas that can be produced from the given reactants is 0.469 grams, not 0.729 grams.
To calculate the mass of hydrogen gas produced in the given chemical reaction, you need to use the stoichiometry of the reaction. The stoichiometry tells you the molar ratio between the reactants and products.
First, let's calculate the number of moles of SrH2 and H2O using their respective molar masses. The molar mass of SrH2 is 105.62 g/mol, and the molar mass of H2O is 18.02 g/mol.
Number of moles of SrH2 = Mass of SrH2 / Molar mass of SrH2
= 5.22 g / 105.62 g/mol
≈ 0.0494 mol
Number of moles of H2O = Mass of H2O / Molar mass of H2O
= 4.35 g / 18.02 g/mol
≈ 0.241 mol
Next, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, thus limiting the amount of product that can be formed. To find the limiting reactant, compare the moles of the reactants using the coefficients in the balanced equation.
The balanced equation is:
SrH2 (s) + 2H2O (l) → Sr(OH)2 (s) + 2H2 (g)
The molar ratio between SrH2 and H2 is 1:2, which means that for every 1 mole of SrH2, 2 moles of H2 are produced. Therefore, the number of moles of hydrogen gas that can be produced from SrH2 is 2 times the number of moles of SrH2.
Number of moles of H2 (from SrH2) = 2 * Number of moles of SrH2
= 2 * 0.0494 mol
≈ 0.0988 mol
The molar ratio between H2O and H2 is 2:2, which means that for every 2 moles of H2O, 2 moles of H2 are produced. Therefore, the number of moles of hydrogen gas that can be produced from H2O is equivalent to the number of moles of H2O.
Number of moles of H2 (from H2O) = Number of moles of H2O
= 0.241 mol
Comparing the number of moles of hydrogen gas produced from SrH2 and H2O, we see that the H2 produced from SrH2 is smaller, indicating that SrH2 is the limiting reactant.
Now, we can calculate the mass of hydrogen gas produced from SrH2 using its molar mass, which is 2.02 g/mol.
Mass of H2 (from SrH2) = Number of moles of H2 (from SrH2) * Molar mass of H2
= 0.0988 mol * 2.02 g/mol
≈ 0.199 grams (rounded to three decimal places)
Therefore, the mass of hydrogen gas that can be prepared from 5.22 grams of SrH2 and 4.35 grams of H2O is approximately 0.199 grams.