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2 years ago

How many grams of sodium phosphate are formed by the reaction of 10 g of sodium with phosphoric acid?

Chemistry

1 answer:

Norma-Jean [14]2 years ago

8 0

Answer:

23.8g of sodium phosphate are formed

Explanation:

Based on the reaction of sodium, Na, with phosphoric acid, H₃PO₄:

3Na + H₃PO₄ → Na₃PO₄ + 3/2 H₂

3 moles of sodium produce 1 mole of sodium phosphate

To solve this question we must find the moles of sodium in 10g. Using the chemical reaction we can find the moles -And the mass- of sodium phosphate produced, as follows:

Moles Na -Molar mass: 22.99g/mol-

10g * (1mol / 22.99g) = 0.435 moles Na

Moles Na₃PO₄:

0.435 moles Na * (1mol Na₃PO₄ / 3mol Na) = 0.145 moles Na₃PO₄

Mass Na₃PO₄ -Molar mass: 163.94g/mol-

0.145 moles Na₃PO₄ * (163.94g/mol) =

<h3>23.8g of sodium phosphate are formed</h3>

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If 17.8 grams of KOH dissolve in enough water to make a 198-gram solution, what is the concentration in percent by mass?

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Solvn = 198 g

% = 17.8 / 198

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3 years ago

Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

Brums [2.3K]

Answer:1) Volume of $AgNO_3$ required is 55.98 mL.

2) 0.62577 grams of $Ag_3PO_4$ is produced.

Explanation:

$3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)$

1) Molarity of $AgNO_3,M_1=0.225 M$

Volume of $AgNO_3.V_1=?$

Molarity of $Na_3PO_4,M_2=0.135 M$

Volume of $Na_3PO_4,V_2=31.1 mL=0.0311 L$

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

$\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles$

According to reaction, 1 mole of $Na_3PO_4$ reacts with 3 mole of $AgNO_3$ , then, 0.0041985 moles of $Na_3PO_4$ will react with:

$\frac{3}{1}\times 0.0041985$ moles of $AgNO_3$ that is 0.0125955 moles.

$M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}$

$V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL$

Volume of $AgNO_3$ required is 55.98 mL.

$Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

Number of moles of $AgNO_3=0.195\times 0.023 L=0.004485 moles$

According to reaction, 3 moles of $AgNO_3$ gives 1 mole of $Ag_3PO_4$ , then 0.004485 moles of $AgNO_3$ will give: $\frac{1}{3}\times 0.004485$ moles of $Ag_3PO_4$ that is 0.001495 moles.

Mass of $Ag_3PO_4$ =

Moles of $Ag_3PO_4$ × Molar Mass of $Ag_3PO_4$

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of $Ag_3PO_4$ is produced.

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Ostrovityanka [42]

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