If this is a true or false question this would be true
Answer:
99.24%.
Explanation:
- NaCl reacted with AgNO₃ as in the balanced equation:
<em>NaCl + AgNO₃ → AgCl(↓) + NaNO₃,</em>
1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.
- We need to calculate the no. of moles of AgCl produced:
no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.
- Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:
<em>using cross multiplication:</em>
1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.
∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.
- Now, we can get the mass of puree NaCl in the sample:
mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.
∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.
Answer : The mass of nitric acid is, 214.234 grams.
Solution : Given,
Moles of nitric acid = 3.4 moles
Molar mass of nitric acid = 63.01 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of nitric acid.
Therefore, the mass of nitric acid is, 214.234 grams.
Answer:
Kc = 12.58
Explanation:
Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3
Kc = (0.052441)(0.10513)/(0.002385)(0.18373)
Kc = 0.0005513/0.000438
Kc = 12.58
Hope that helps!!