Answer:
51.53 grams .
Explanation:
Na₃PO₄ ⇄ 3Na⁺¹ + PO₄⁻³ .
1 mole 3 mole
725 mL of 1.3 M Na⁺ ions
= .725 x 1.3 moles of Na⁺ ions
= .9425 moles
3 mole of Na⁺ is formed by 1 mole of Na₃PO₄
.9425 mole of Na⁺ is formed by .9425/3 mole of Na₃PO₄
Na₃PO₄ needed = .9425/3 moles = .3142 moles
Molecular weight of Na₃PO₄ = 164
grams of Na₃PO₄ needed = .3142 x 164 = 51.53 grams .
The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
The arrangement in space and the interatomic distances and angles of the atoms in crystals, usually determined by x-ray diffraction measurements
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