a) Step 1 - We need to balance the equation:
MO2 (s) 4 HCl (aq) -> MCl2 (aq) + Cl2 (g) + 2 H2O(l)
Reactant side:
M - 1
O - 2
H - 4
Cl - 4
Product side
M - 1
O - 2
H - 4
Cl - 4
Step 2 - We need to transform 0.20 grams of MO2 into moles using the following formula: mole = mass/molar mass
mass = 0.20 g
molar mass = 87 g/mol
mole = 0.002299
0.002299 moles of MO2
Step 3 - let's transform mL into L and then transform 0.15 M = 0.15 mol/L into moles.
1000 mL = 1 L
25 mL = 0.025 L
0.15 moles --- 1 L
x --- 0.025 L
x = 0.00375 moles
0.00375 moles of HCl
Step 4 - Let's see the proportion of the equation and compare with the real value of moles
1 mole of MO2 reacts with 4 moles of HCl
So:
1 MO2 --- 4 HCl
0.002299 MO2 --- x
x = 0.00919 moles of HCl
1 MO2 --- 4 HCl
x MO2 --- 0.00375
x = 0.0009375 moles of MO2
We should have 0.00919 moles of HCl reacting with 0.002299 moles of MO2, but we have just 0.00375. It means that the limiting reactant is HCl.
a) HCl is the limiting reactant.
b) Step 1 - We first need to find the atomic mass of M. For this, we know that molar mass of O is 16 and the molar mass of MO2 is 87.
So:
M + (2x16) = 87
M = 87 - 32
M = 55 g/mol
molar mass of M is 55 g/mol.
Step 2 - Now we need to find the molar mass of MCl2:
(1x55) + (35.45 x 2) = 125.9 g/mol
Step 3 - Now we use the equation proportion to find the quantity in moles of MCl2 produced:
1 mole of MO2 produces 1 mole of MCl2
0.0009375 moles of MO2 produces 0.0009375 moles of MCl2
Step 4 - Transform moles of MCl2 into grams:
125.9 g --- 1 mol
x --- 0.0009375 moles of MCl2
x = 0.118 grams
b) Mass of MCl2 produced is 0.118 grams.
c) To calculate the percentage yield we use the following formula:
Percent yield = (actual yield / theoretical yield) x 100%
actual yield = 0.078
theoretical yield = 0.118
Percent yield = (0.078 / 0.118) x 100%
c) Percent yield = 66%
Answer:
a) HCl is the limiting reactant.
b) Mass of MCl2 produced is 0.118 grams.
c) Percent yield = 66%