Fe2o3+2al=al2o3+2fe a welder has 1.873 × 10^2 g fe2o3 and 94.51 g al in his welding kit. which reactant will he run out of first? (options: fe2o3 or 's fe2o3). how much of this reactant should he order to make sure he runs out of both reactants at the same time? he should order of the limiting reactant.

a)

b) 92.1 g

Explanation:

To calculate the moles, we use the equation:

a) moles of

b) moles of

According to stoichiometry :

1 mole of require 2 moles of

Thus 1.173 require=  of

Thus is the limiting reagent as it limits the formation of product. Thus will run out first.

is the excess reagent as (3.500-2.346)= 1.154 moles are left unreacted.

2 moles of require 1 mole of

3.500 moles of require= of

Moles of  required = (1.750-1.173) = 0.577

Mass of

Thus he should order 92.1 g of the limiting reactant.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying