First you need to find the balanced chemical formula.
CuCl₂+2NaNO₃→2NaCl+Cu(NO₃)₂
Then you can find out how much NaCl 31.0g of CuCl₂ should produce using stoichiometry. Divide 31.0g by the molar mass of CuCl₂ (134.446g/mol) to get 0.2306mol CuCl₂. Than multiply 0.2306mol CuCl₂ by 2 to get 0.4612mol NaCl. Than multiply 0.4612mol by the molar mass of NaCl (58.45g/mol) to get 26.95g of NaCl.
that means that 100% yield would give you 26.95g of NaCl so to find percent yield divide 21.2 by 26.95 to get 0.7867 which is 78.7% yield
Therefore the answer is 78.7% yield.
I hope this helps. Let me know if anything is unclear
Answer:
0.43×10²³ atoms
Explanation:
Given data:
Mass of LiBrO₂ = 4.28 g
Number of atoms of oxygen = ?
Solution:
Number of moles = mass/molar mass
Number of moles = 4.28 g/ 118.84 g/mol
Number of moles = 0.036 mol
We can see 1 mole of LiBrO₂ contain 2 mole of oxygen atm.
0.036 mol × 2 = 0.072 mol
1 mole contain 6.022×10²³ atoms
0.072 mol × 6.022×10²³ atoms / 1mol
0.43×10²³ atoms
Answer: Glucose is a simple sugar with six carbon atoms and one aldehyde group.
Explanation: hope this helps you
Answer:
Explanation:
The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT
ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution
Kb for water = .51°C /m
moles of sugar = 16.90 / 342.3
= .04937 moles
m = moles of sugar / kg of water
= .04937 / .04090
= 1.207
ΔT = Kb x m
= .51 x 1.207
= .62°C .
So , boiling point of water = 100.62°C .
The balanced reaction: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
We first convert volume of C2H6 to no. of moles. We use the conditions at STP where 1 mol = 22.4 L thus,
Moles C2H6 = 16.4 L/ 22.4 L =0.7321 mol
In order to determine the limiting reagent, we look at the given amounts of the reactants.
0.7321 mol C2H6 (7/2 mol O2 / 1 mol C2H6) = 2.562 mol O2
From the given amounts of the reactants, we can say that O2 is the limiting reactant since we need 2.562 mol O2 to completely react the given amount of C2H6. The excess reagent is C2H6
To calculate for the amount of products and excess reactants:
0.980 mol O2 (2 mol CO2 / (7/2 mol O2)) = 0.56 mol CO2 (22.4 L / 1 mol ) =12.544 L CO2
<span>0.980 mol O2 (1 mol C2H6 / (7/2 mol O2)) = 0.28 mol C2H6
Excess C2H6 = 0.7321 mol - 0.28 mol C2H6 = 0.4521 mol C2H6
We then use the molecular weight of C2H6 to convert the excess amount to grams.
0.4521 mol C2H6 (30.08 g C2H6 / 1 mol C2H6) = <span>13.60 g C2H6
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<span>Since the limiting reagent is O2 there will be no oxygen atoms that will be left after the reaction.</span>
