Write the complete balanced equation for the decomposition of AI(CIO3)3.

Sn + 2 H2SO4 → SnSO4 + SO2 + 2 H2O If 219.65 grams of SnSO4 are produced, how many moles of H2SO4 were reacted?

anygoal [31]

Answer:

2.05moles

Explanation:

The balanced chemical equation in this question is as follows;

Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O

Based on the above equation, 2 moles of H2SO4 reacted to produce 1 mole of SnSO4

However, the mass of SnSO4 produced is 219.65 grams. Using mole = mass/molar mass, we can find the number of moles of SnSO4 produced.

Molar mass of SnSO4 where Sn = 118.7, S = 32, O = 16

= 118.7 + 32 + 16(4)

= 150.7 + 64

= 214.7g/mol

mole = 219.65/214.7

mole = 1.023mol

Therefore, if 2 moles of H2SO4 reacted to produce 1 mole of SnSO4

1.023 mol of SnSO4 produced will cause: 1.023 × 2/1

= 2.046moles of H2SO4 to react.

8 0

3 years ago

How does melting order relate to melting point?

KatRina [158]

Answer:

I think A.

Explanation:I say A because of the substance melting the quicking does have the highest melting point because its the highest.

5 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

3 years ago

I just need the right answer plz

astraxan [27]

Answer:

-0.5

Explanation:

5 0

3 years ago

ubstance A is a nonpolar liquid and has only dispersion forces among its constituent particles. Substance B is also a nonpolar l

LenaWriter [7]

Answer:

Their particles exhibit the same type of intermolecular interaction

Explanation:

In chemistry, we commonly say that 'like dissolves like'. This implies that polar solvents dissolves polar solutes while nonpolar solvents dissolve nonpolar solutes.

This phenomenon of 'like dissolves like' is possible because, the dissolution of one substance in another involves intermolecular interaction between the solute and solvent molecules.

If the molecules of solute and solvent are both nonpolar and have about the same magnitude of intermolecular (dispersion) forces, interaction between the both molecules is significant hence the solute dissolves completely in the solvent.

6 0

3 years ago