6.75 moles of H₂
Explanation:
We have the following balanced chemical reaction:
Mg + 2 HCl → MgCl₂ + H₂
From the chemical reaction we deduce that if 1 mole of Mg is reating with 2 moles of HCl then 8.30 moles of Mg is reaction with 16.60 moles of HCl, quantity which is over our available 13.5 moles of HCl. The limiting reactant is HCl.
Knowing this we devise the following reasoning:
if 2 moles of HCl produces 1 mole of H₂
then 13.5 moles of HCl produces X moles of H₂
X = (13.5 × 1) / 2 = 6.75 moles of H₂
Learn more about:
balancing chemical equations
brainly.com/question/13898428
#learnwithBrainly
Answer:
La masa de óxido de carbono iv formado es 44 g.
Explanation:
En esta pregunta, se nos pide calcular la masa de óxido de carbono iv formado a partir de la reacción de masas dadas de carbono y oxígeno.
En primer lugar, necesitamos escribir una ecuación química equilibrada.
C + O2 → CO2
De la ecuación, 1 mol de carbono reaccionó con 1 mol de oxígeno para dar 1 mol de óxido de carbono iv.
Ahora, si marca las masas en la pregunta, verá que corresponde a la masa atómica y la masa molar de la molécula de carbono y oxígeno, respectivamente. ¿Qué indica esto?
Como tenemos una relación molar de 1: 1 en todo momento, lo que esto significa es que la masa de óxido de carbono iv producida también es la misma que la masa molar de óxido de carbono iv.
Por lo tanto, procedemos a calcular la masa molar de óxido de carbono iv Esto es igual a 12 + 2 (16) = 12 + 32 = 44 g Por lo tanto, la masa de óxido de carbono iv formado es 44 g
Answer:
11.3 g of 
are produced from 36.0 g of 
Explanation:
1. The balanced chemical equation is the following:
2. Use the molar mass of the , the molar mass of the and the stoichiometry of the balanced chemical reaction to find how many grams of are produced:
Molar mass = 18
Molar mass = 17
Therefore 11.3 g of are produced from 36.0 g of
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;
= 0.8
The rate-out
= 
= 
We can say that:
where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12
Integration of the above linear equation =
so we have:
![\frac{d}{dt}[e^{\frac{1}{3}t}A]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%5Be%5E%7B%5Cfrac%7B1%7D%7B3%7Dt%7DA%5D)
∴ 
Since A(0) = 12
Then;
Hence;
∴ the concentration at 10 minutes is ;
= 
%
= 0.0456667 %
= 0.046% to three decimal places
