Asked by Monica
A solution of trioxonitrate(v)acid contains 0.63g in 100cm³. 28.0cm³of this solution neutralized 25cm³of sodiumtrioxocarbonate(iv) solution. Calculate the molar concentration and mass concentration of sodiumtrioxocarbonate(iv)
Answers
Answered by
DrBob222
HNO3 = 0.63 g/100 c
(HNO3) = mols/L = g/molar mass/L = 0.63/63/0.100L = 0.1 M
2HNO3 + Na2CO3 ==> 2NaNO3 + H2O + CO2
millimols HNO3 = mL x M = 28.0 mL x 0.1 M = 2.80
2.8 mmols HNO3 will neutralize 2.80 x (1 mol Na2CO3/2 mol HNO3) = 1.40 mmols Na2CO3 .
The (Na2CO3) = mols/L = millimoles/mL = 14.0/25 = ? M
(HNO3) = mols/L = g/molar mass/L = 0.63/63/0.100L = 0.1 M
2HNO3 + Na2CO3 ==> 2NaNO3 + H2O + CO2
millimols HNO3 = mL x M = 28.0 mL x 0.1 M = 2.80
2.8 mmols HNO3 will neutralize 2.80 x (1 mol Na2CO3/2 mol HNO3) = 1.40 mmols Na2CO3 .
The (Na2CO3) = mols/L = millimoles/mL = 14.0/25 = ? M
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