All categories

3 years ago

How many moles of Na3PO4 would be required to react with 1.0 mol of AgNO3?

Chemistry

2 answers:

Solnce55 [7]3 years ago

6 0

1/3 mol or .3 repeating, I believe.

Tema [17]3 years ago

5 0

<u>Answer:</u> The number of moles of sodium phosphate required is 0.33 moles

<u>Explanation:</u>

We are given:

Moles of silver nitrate = 1 mole

The chemical equation for the reaction of sodium phosphate and silver nitrate follows:

$Na_3PO_4+3AgNO_3\rightarrow 3NaNO_3+Ag_3PO_4$

By Stoichiometry of the reaction:

3 moles of silver nitrate reacts with 1 mole of sodium phosphate.

So, 1.0 moles of silver nitrate will react with = $\frac{1}{3}\times 1.0=0.33mol$ of sodium phosphate.

Hence, the number of moles of sodium phosphate required is 0.33 moles

You might be interested in

How much does the Earth weigh?

ELEN [110]

5.972 × 10^24 kg

hope it helps

6 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

<u>Answer:</u> The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

<u>Initial:</u> 1.30 1.65

<u>At eqllm:</u> 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

2 years ago

Read 2 more answers

Which is NOT true of ionic bonding?

WARRIOR [948]

Answer:

D is wrong

Explanation:

it doesn't involves any kind of sharing of electrons

6 0

2 years ago

Scorpion4ik [409]

Answer:

0.24M

Explanation:

The equation for the reaction is given below:

H2SO4 + 2KOH → K2SO4 + 2H2O

From the equation above, we obtained the following information:

nA (mole of acid) = 1

nB (mole of base) = 2

Data obtained from the question include:

Va (volume of the acid) = 12mL

Ca (concentration of the acid) =?

Vb (volume of the base) = 36mL

Cb (concentration of the base) = 0.16 M

The Ca (concentration of the acid) can be obtained as follow:

CaVa/CbVb = nA/nB

Ca x 12 / 0.16 x 36 = 1 /2

Cross multiply to express in linear form as shown below:

Ca x 12 x 2 = 0.16 x 36

Divide both side by 12 x 2

Ca = 0.16 x 36/ 12 x 2

Ca = 0.24M

Therefore, the concentration of the acid is 0.24M

6 0

3 years ago

Select all the correct answers.

Lera25 [3.4K]

The correct answers :

It increases with a decrease in the concentration of H₂(g).

It decreases with an increase in the concentration of S₂(g).

It decreases with an increase in the concentration of H₂(g).

<h3 /><h3>Further explanation</h3>

Forward reaction : rate to form product

In equilibrium :

The product decreases ⇒ system will move from left to right(forward reaction)

The product increases ⇒ system will move from right to left(reverse reaction)

6 0

2 years ago