Calculate the number of grams of solid aluminium chloride that will form when a mixture containing 0.150 g of aluminum powder and 1.00 g of chlorine gas is allowed to react. 2al (s) + 3cl2 (g) > 2alcl3 (s)

0.630 g

Step-by-step explanation:  

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, lets assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:    26.98   70.91      133.34  

           2Al   +   3Cl₂ ⟶ 2AlCl₃

m/g:  0.150       1.00

Step 1.  Calculate the moles of each reactant  

Moles of Al   = 0.150 g × 1 mol/26.98 g  = 0.005 560 mol

Moles of Cl₂ =  1.00  g  × 1 mol/70.91  g  = 0.014 10     mol

Step 2. Identify the limiting reactant

Calculate the moles of AlCl₃ we can obtain from each reactant.

From Al:  

The molar ratio of AlCl₃:Al is 2:2

Moles of AlCl₃ = 0.005 560 × 2/2  

Moles of AlCl₃ = 0.005 560 mol AlCl₃

From Cl₂:  

The molar ratio of AlCl₃: Cl₂ is 2:3.  

Moles of AlCl₃ = 0.014 10 × 2/3  

Moles of AlCl₃ = 0.009 401 mol AlCl₃  

Al is the limiting reactant because it gives the smaller amount of AlCl₃.  

Step 4. Calculate the theoretical yield.

Theor. yield = 0.005 560 mol AlCl₃ × 133.34 g AlCl₃/1 mol AlCl₃

Theor. yield = 0.630 g AlCl₃

the leaf is an organ of a typical plant, and the main functions are to complete photosynthesis and allow the plant to exchange gas, just like how we have the respiratory system. leaves have the stomata, which collects maintains co2 and o2 levels. this is necessary for cell metabolism. the leaf also collects sunlight using chloroplasts. succinctly, leaves own a pivotal role in sustaining plant life.

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