Answer: 60 ml of the 65% mixture will you need to add to obtain the desired solution.
Explanation:
According to the dilution law,
[tex]C_1V_1+C_2V_2=C_3V_3[/tex]
where,
= concentration of given alcohol solution = 25 %
[tex]V_1[/tex] = volume of given alcohol solution = 100 ml
[tex]C_2[/tex] = concentration of another alcohol solution=65 %
[tex]V_2[/tex] = volume of another acid solution= x ml
[tex]C_3[/tex] = concentration of resulting alcohol solution = 40 %
[tex]V_3[/tex] = volume of resulting acid solution = (100+x)
[tex]25\times 100+65\times x=40\times (100+x)[/tex]
[tex]x=60ml[/tex]
Thus 60 ml of the 65% mixture will you need to add to obtain the desired solution.
How much HCl is produced from the reaction of an excess of HSbCl4 with 3 moles H2S in the following reaction? HSbCl4 + H2S → Sb2S3 + HCl (Remember to balance the equation.)
Answer:
We will produce 8.0 moles of HCl , this is 291.7 grams HCl
Explanation:
Step 1: Data given
Number moles of H2S = 3.0 moles
Step 2: The balanced equation
2HSbCl4 + 3H2S → Sb2S3 + 8HCl
Step 3: Calculate moles HCl
For 2 moles HSbCl4 we need 3 moles H2S to produce 1mol Sb2S3 and 8 moles HCl
For 3.0 moles H2S we'll have 8.0 moles HCl
Step 4: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
Mass HCl = 8.0 moles * 36.46 g/mol
Mass HCl = 291.7 grams
We will produce 8.0 moles of HCl , this is 291.7 grams HCl
From the balanced equation, it is determined that 2 moles of HCl are produced from 1 mole of HSbCl4. Therefore, 6 moles of HCl will be produced from the reaction of an excess of HSbCl4 with 3 moles of H2S.
Explanation:The balanced equation for the reaction is:
HSbCl4 + H2S → Sb2S3 + 2HCl
The mole ratio between HSbCl4 and HCl is 1:2, which means that for every 1 mole of HSbCl4, 2 moles of HCl are produced.
Since there is an excess of HSbCl4, we can assume that all 3 moles of H2S will react.
Therefore, the number of moles of HCl produced will be:
(3 moles H2S) x (2 moles HCl/1 mole HSbCl4) = 6 moles HCl