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3 years ago

What is the formula for potassium citrate and how do you find it?

Chemistry

2 answers:

oee [108]3 years ago

7 0

Answer:

Potassium citrate (also known as tripotassium citrate) is a potassium salt of citric acid with the molecular formula K3C6H5O7.

Chemical formula: K3C6H5O7

Density: 1.98 g/cm3

Melting point: 180 °C (356 °F; 453 K)

Boiling point: 230 °C (446 °F; 503 K)

Explanation:

Shkiper50 [21]3 years ago

6 0

Answer:

K3c6h5o7

Explanation:

the formula of potassium citrate is

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Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how

Afina-wow [57]

Answer:

11.3 g of $NH_{3}$ are produced from 36.0 g of $H_{2}O$

Explanation:

1. The balanced chemical equation is the following:

$Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)$

2. Use the molar mass of the $H_{2}O$ , the molar mass of the $NH_{3}$ and the stoichiometry of the balanced chemical reaction to find how many grams of $NH_{3}$ are produced:

Molar mass $H_{2}O$ = 18 $\frac{g}{mol}$

Molar mass $NH_{3}$ = 17 $\frac{g}{mol}$

$36.0gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesNH_{3}}{6molesH_{2}O}*\frac{17gNH_{3}}{1molNH_{3}}=11.3gNH_{3}$

Therefore 11.3 g of $NH_{3}$ are produced from 36.0 g of $H_{2}O$

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