<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
Answers:
(a) 30.55 °C
(b) 298 K and 77°F
(c) 204.44 °C and 477.44 K
(d) -320.8 °F and -196 °C
Explanation:
Converting °C into °F;
°F = °C × 1.8 + 32
Converting °F into °C;
°C = °F - 32 ÷ 1,8
Converting °C into K;
K = °C + 273
Converting K into °C;
°C = K - 273
Answer:
10.85 g of water
Explanation:
First we write the balanced chemical equation
Then we calculate the number of moles of nitric acid produced
n(HNO3) = 
According to the balanced equation, water needed in moles is always half the number of moles of HNO3 produced. So since we will produce 1.2044 mol of HNO3, we will need 0.6022 mol of water. Now to calculate what mass that is:
mass(water)=number of moles*molar mass=0.6022mol*18.02g/mol=10.85g
Answer:
for the given reaction is -238.7 kJ
Explanation:
The given reaction can be written as summation of three elementary steps such as:
---------------------------------------------------------------------------------------------------
