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shtirl [24]
3 years ago
6

65.5g of copper reacts with __g of oxygen to produce 81g copper(I) oxide

Chemistry
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

grams O₂(g) ≅ 15.9 g O₂(g) (3 sig. figs.)  

Explanation:

2Cu + O₂ => 2CuO

Convert given to moles, solve by equation rxn ratio then convert to grams.

Moles CuO formed = 81.0 g / 81.5 g·mol⁻¹ = 0.994 mol CuO

0.994 mol CuO requires 0.994 mol Cu = 0.994 mol Cu x 64 g·mol⁻¹ = 63.6 g Cu => from this, the amount of O₂ needed = 1/2(0.994 mol) O₂(g) = 0.497 mol O₂(g) = (0.497 mol O₂(g))(32 g·mol⁻¹) = 15.904 g O₂(g) ≅ 15.9 g O₂(g) (3 sig. figs.)                

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6 0
2 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

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

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