when iron and oxygen the reactin form Fe2O3 balance equation for this is <span>4Fe + 3O2---> 2Fe2O3
9moles of O2 are needed to produce 6mol of Fe2O3 since the ratio of oxygen to iron(II)oxide is 3:2
hope its help</span>

6 0

3 years ago

What mass of solute must be used to prepare 500ml of 0.100M aqueous sodium borate Na2B4O7 from solid hydrated sodium borate Na2B

motikmotik

0.000132 g of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O)

Explanation:

First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume (L)

number of moles of Na₂B₄O₇ = 0.1 × 0.5 = 0.05 moles

We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.

Now to find the mass of hydrated sodium borate we use the following formula:

number of moles = mass / molar weight

mass = number of moles × molar weight

mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g

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molar concentration

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4 0

2 years ago

Consider the decomposition of hydrogen peroxide reaction. If the rate of decomposition of hydrogen peroxide at a particular temp

Natasha2012 [34]

Answer:

0.049 mol/L.s

Explanation:

The decomposition of hydrogen peroxide is:

$H_2O_2 \to H_2O + \dfrac{1}{2}O_2$

$Rate = -\dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{ 2 \Delta [H_2O_2]}{\Delta t}$

The rate of decomposition reaction = the rate of formation of $H_2O$ = 0.098 mol/L.s

∴

Rate of formation of $O_2$ $= \dfrac{ rate \ of \ reaction }{2}$

$= \dfrac{ 0.098 }{2}$

= 0.049 mol/L.s

8 0

2 years ago