The alcohol concentration of the mixed solution is 20%
Simplification :
Based on the given condition, formulate :
35% ×0.40 + 0.6 ×10% ÷{ 0.4+0.6}
Calculate the product :
Calculate the sum or difference : 
Any fraction with denominator 1 is equal to numerator : 0.2
Multiply a number to both numerator, denominator : 0.2 ×
Calculate the product or quotient : 
A fraction with denominator equals to 100 to a percentage 20%.
How do you find the concentration of a mixed solution?
In general when your are mixing two different concentrations together first calculate number of moles for each solution (n=CV ,V-in liter) then add them together it will be total moles,then concentration of mixture will be = total moles / total volume(liter).
Learn more about concentration of alcohol :
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Mass=density·volume. The density is 2.70g/mL and the volume is 353mL. So you would multiply 2.70g/mL by 353mL which will give you 953.1g. Hope that helps :)
A) Limiting reactant
You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)
a) Molar ratios:
6 mol HF : 1 mol SiO2 : 1 mol H2SiF6
2) Molar masses:
Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol
=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol
3) convert data in grams to moles
21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2
70.5 g HF / 20 g/mol = 3.525 mol HF
4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.
6 mol HF / 1mol SiO2 < 3.525 mol HF / 0.35 mol SiO2 ≈ 10
=> There is more HF than the needed to react with 0.35mol of SiO2 =>
SiO2 is the limiting reactant (HF is in excess)
b) Mass of excess reactant.
1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted
0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF
2.1 mol HF * 20 g/mol = 42 gram of HF
2) Subtract the quantity that reacted from the original quantity:
70.5 g - 42 g = 28.5 g of HF in excess
c) Theoretical yield of H2SiF6
1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6
Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams
d) % yield
% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%
