the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g
The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn
3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2
the formula is n= mass/M so, now substituting values
m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3
so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g
so mass of aluminum oxide obtained = 1.36g
To learn more about Mass:
brainly.com/question/19694949
#SPJ4
This may help you
First write and balance the equation, being:
CaCO3 - CaO + CO2
Then, using the periodic table, find the molecular masses of CaCO3 and of CaO, finding their ratio. That will be 100g:56g or 0.1kg:0.056kg. Since you have 4.7kg of CaCO3, it corresponds to Xkg of CaO. Making x the subject, it should be X= 4.7*0.056/100=0,002632
43.8 has 3 significant figures and 1 decimal.
<h3 /><h3>What are significant figures?</h3>
The term significant figures refer to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation.
All zeros that occur between any two non-zero digits are significant. For example, 108.0097 contains seven significant digits. All zeros that are on the right of a decimal point and also to the left of a non-zero digit are never significant. For example, 0.00798 contained three significant digits.
Hence, 43.8 has 3 significant figures and 1 decimal.
Learn more about significant figures here:
brainly.com/question/14359464
#SPJ1
Answer:
Are you in flvs, if so im prettyb sure if yo look on page 3 of lesson 1.04 it tells you the answer.
Explanation:
