If 56.8 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 554 mg sample of na2so4 (forming baso4), what is
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Answer:
0.15g
Explanation:
Given parameters:
Number of molecules of water = 1.2 x 10²¹ molecules
Unknown:
Mass of SnO₂ = ?
Solution:
To solve this problem, we have to work from the known to the unknown specie;
SnO₂ + 2H₂ → Sn + 2H₂O
Ensure that the equation given is balanced;
Now,
the known species is water;
6.02 x 10²³ molecules of water = 1 mole
1.2 x 10²¹ molecules of water = = 0.2 x 10⁻²moles
Number of moles of water = 0.002moles
From the balanced chemical equation:
2 mole of water is produced from 1 mole of SnO₂
0.002 moles of water will be produced from = 0.001moles
To find the mass;
Mass = number of moles x molar mass
Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol
Mass = 0.001 x 150.7 = 0.15g