Answer:
0.09375M
Explanation:
There are two methods of going about this, either we use dilution formula (easiest and fastest) or we solve through molarity.
Using dilution formula,
C2 (H2SO4) = ?
C1 (NaOH) = 0.25M
V2 (H2SO4) = 20cm³
V1 (NaOH) = 15cm³
However we can solve using molarity method
Equation of reaction =
2NaOH + H2SO4 ====》 Na2SO4 + 2H2O
O.25M of NaOH = 1000cm³
X moles = 15cm³
X = (0.25 * 15) / 1000
X = 0.00375 moles is present in 15cm³ of NaOH
From equation of reaction,
2 moles of NaOH requires 1 mole of H2SO4
Therefore
0.00375 / 2 = 0.001875 moles is present in H2SO4
From the reaction,
0.00187 moles of H2SO4 = 20 cm³
X moles = 1000cm³
X = (0.00187*1000) / 20 = 0.09375M
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The answer is
You must make sure each reactant has the same amount of subscripts on both sides of the equation.
Answer:
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a) (NH4)2SO4 --- 1 mole of it contains 2 moles of N, 8 moles of H, 1 mole of S, and 4 moles of O.
MM = (2 moles N x 14.0 g/mole) + (8 moles H x 1.01 g/mole) + (1 mole S x 32.1 g/mole) + (4 moles O x 16.0 g/mole) = 132 g/mole.
6.60 g (NH4)2SO4 x (1 mole (NH4)2SO4 / 132 g (NH4)2SO4) = 0.0500 moles (NH4)2SO4
b) The molar mass for Ca(OH)2 = 74.0 g/mole, calculated like (NH4)2SO4 above.
4.5 kg Ca(OH)2 x (1000 g / 1 kg) x (1 mole Ca(OH)2 / 74.0 g Ca(OH)2) = 60.8 moles Ca(OH)2
