Its B. your welcome star me? pls????
They all have their different measurements like the volume of a liquid can be measured in a lab with a beaker.
Respuesta:
199.5 g
Explicación:
Paso 1: Escribir la reacción balanceada
2 Al + 3 H₂SO₄ ⇒ Al₂(SO₄)₃ + 3 H₂
Paso 2: Calcular la masa pura de 50 g de Al
Aluminio tiene 10% de impurezas, es decir, 10% de 50 g = 5 g. Luego, tiene 50 g - 5 g = 45 g de Al puro.
Paso 3: Calcular la masa teórica de Al₂(SO₄)₃ obtenida a partir de 45 g de Al
La relación de masas de Al₂(SO₄)₃ a Al es 342:54.
45 g Al × 342 g Al₂(SO₄)₃/54 g Al = 285 g Al₂(SO₄)₃
Paso 4: Calcular la masa real de Al₂(SO₄)₃ obtenida
El rendimiento de la reacción es de 70%.
285 g × 70% = 199.5 g
Honestly? Try putting the elements on each side of the reaction and count how many you have and how many you need to balance it for example:
CH4 + O2 => H2O+ CO2
Reactants:
1 C
4 H
2 O
Products:
1 C
2 H
3 O
So basically what you do now...you see that you have one carbon on each side but 4 hydrogen on recatants side and 2 hydrogen on the products side, so you need to multiply the water molecule by 2 to get the 4 hydrogens. Now because you multiplied (put coefficient 2 in front of it) the oxygen in h20 is now 2 oxygens right? So now you have 4 oxygens on the product side and 2 on the reactant side so in the reactant side, y più multiply it by 2 to get 4 oxygens.
C10H15O5
C2=24.0g/mol
H3= 3.0g/mol
0= 16.0g/mol
__________
=43.0g/mol
215.2g/mol/43.0g/mol=aproximately 5
multiply the empirical formula by 5 to get the molecular formula
C2H3O*5=C10H15O5