Answer: 50. 4g
Explanation:
First calculate number of moles of aluminium in 38.8g
Moles = 38.8g/ 26.982mol/g
= 1.44mol
By looking at the balance equation you can see that 4 moles of aluminium produce 2 moles of aluminium oxide.
4 = 2
1.4 = x
Find the value of x
x= (1.4×2)/4= 0.72 mol
0.72 moles of aluminium oxide are produced from 38.8g of aluminium
Now find the mass of aluminium produced.
Mass = moles × molar mass
= 0.72mol × 69.93 mol/g
= 50.4g
Answer:
Explanation: In my opinion law is stated as fact because it is passes through and approved by many peoples but theory is an idea someone has and is not approved and is set on a base of beliefs.
The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.
Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.
We recognise that
h * 2 = PT - P * h * 20 = 1atm - 0.037atm
= 0.963 atm
1 * h * 2 = Ph * 2V / R * T
= 0.963 atm x 0.520 L / 0.0821 L atm/
molK * 301
= 0.02 mol h2
= 0.02molZn
So 0.02 mol Zn x 65.39 g/mol
= 1.33 g Zn
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B
step-by-step explanation
