A mixture of copper and tin would be called ________

In what ways do scientists not study the history of Earth's climate?

BARSIC [14]

(A) Studying ice cores

3 0

3 years ago

Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide

Contact [7]

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

According to question

2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

$'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\$

Since 'NO' is the limiting reagent according to this ratio.

According to equation

2 moles NO reacts to form 2 moles NO₂

So, 0.1855 moles NO give = 0.1855 moles of NO₂

Mass of 1 mole NO₂ = 46 g/mole

Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

5 0

3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$