Whitney is making sweet tea. She adds 15 grams of

If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma

Veseljchak [2.6K]

Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

= 29.07 gram of Cu(NO3)2

4 0

3 years ago

When excess dilute hydrochloric acid was added to sodium sulphite 960 of sulphuric (iv) oxide was produced. calculate the mass o

irakobra [83]

The mass of sodium sulfite that was used will be 1,890 grams.

<h3>Stoichiometric problems</h3>

First, the equation of the reaction:

$NaSO_3 + 2HCl --- > NaCl_2 + H_2O + SO_2$

The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.

Mole of 960 grams SO2 = 960/64 = 15 moles

Equivalent mole of sodium sulfite that reacted = 15 moles

Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

3 0

2 years ago

For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper

Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1\text{ and }V_1$ are initial pressure and volume at initial temperature $T_1$ .

$P_2\text{ and }V_2$ are final pressure and volume at initial temperature $T_2$ .

We are given:

$P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V$

$T_1=-25^oC=248.15 K$

$T _2 = 16^oC=289.15 K$

Putting values in above equation, we get:

$\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}$

$P_1=0.225 kPa$

Hence, the new pressure will be 0.225 kPa.

4 0

4 years ago

What is the molarity of a solution in which 7.1 g of sodium sulfate is dissolved in enough water to make 100. mL of solution?

Sidana [21]

Answer:

0.50 M

Explanation:

Given data

Mass of sodium sulfate (solute): 7.1 g

Volume of solution: 100 mL

Step 1: Calculate the moles of the solute

The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

$7.1g \times \frac{1mol}{142.02g} = 0.050mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1 L = 1000 mL.

$100mL \times \frac{1L}{1000mL} =0.100L$

Step 3: Calculate the molarity of the solution

$M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.050mol}{0.100L} =0.50 M$

3 0

3 years ago

Read 2 more answers

A compound is determined to have the empirical formula C2OH4. If the molar mass of the compound is 132 g/mol, determine the mole

8_murik_8 [283]

Answer: The molecular formula will be $C_6O_3H_{12}$

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

Empirical weight of $C_2OH_{4}$ is $12\times 2+16\times 1+1\times 4=44g$

Molecular mass of compound is = 132 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{132}{44}=3$

The molecular formula will be= $3\times C_2OH_4=C_6O_3H_{12}$

5 0

3 years ago