Question

reaction product 3/3
consider: how many grams of sio₂ are produced when 13.8 grams of sio is reacted with 13.8 grams of o₂ according to balanced reaction no. 7?

1. how many grams of the excess reactant was consumed in the reaction?
2. 2co + o₂ → 2co₂
3. 2no + 2co → n₂+2co₂
4. 2geo + o₂ → 2geo₂
5. 2feo + o₂ → 2feo₂
6. 2sno + o₂ → 2sno₂
7. 2no + 2geo → n₂ + 2geo₂
8. 2sio + o₂ → 2sio₂
9. 2no + 2sio → n₂ + 2sio₂
10. 2pbo + o₂ → 2pbo₂
11. 2nio + o₂ → 2nio₂
12. 2no + 2sio → n₂ + 2sio₂
13. 2no + 2sno → n₂+2sno₂

hint: the excess reactant used is governed by the limiting reactant. table: see periodic table for molar mass values.
this problem can be solved by dimensional analysis as the primary method, formula where permitted, or simple math. using ratio and proportion methods is not allowed. be sure to show the formula on homework and exams.
show work here as a file upload in the last question. be sure to report the units for all values when showing your work. keys will be posted for exemplary practice problems.

Explanation:

Step1: Calculate moles of reactants

The molar mass of $SiO$ is approximately $M_{SiO}=28.09 + 16=44.09$ g/mol. The number of moles of $SiO$, $n_{SiO}=\frac{m_{SiO}}{M_{SiO}}=\frac{13.8}{44.09}\approx0.313$ mol. The molar mass of $O_2$ is $M_{O_2}=2\times16 = 32$ g/mol. The number of moles of $O_2$, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{13.8}{32}=0.43125$ mol.

Step2: Determine the limiting reactant

From the balanced - equation $2SiO+O_2
ightarrow2SiO_2$, the mole ratio of $SiO$ to $O_2$ is $2:1$. For 0.313 mol of $SiO$, the moles of $O_2$ required is $n_{O_2 - required}=\frac{0.313}{2}=0.1565$ mol. Since $0.1565\lt0.43125$, $SiO$ is the limiting reactant.

Step3: Calculate moles of $SiO_2$ produced

From the balanced - equation, the mole ratio of $SiO$ to $SiO_2$ is $1:1$. So, the moles of $SiO_2$ produced, $n_{SiO_2}=n_{SiO}=0.313$ mol.

Step4: Calculate mass of $SiO_2$ produced

The molar mass of $SiO_2$ is $M_{SiO_2}=28.09+2\times16 = 60.09$ g/mol. The mass of $SiO_2$ produced, $m_{SiO_2}=n_{SiO_2}\times M_{SiO_2}=0.313\times60.09\approx18.81$ g.

Step5: Calculate moles of $O_2$ consumed

Since the mole ratio of $SiO$ to $O_2$ is $2:1$, the moles of $O_2$ consumed, $n_{O_2 - consumed}=\frac{n_{SiO}}{2}=\frac{0.313}{2}=0.1565$ mol.

Step6: Calculate mass of $O_2$ consumed

The mass of $O_2$ consumed, $m_{O_2 - consumed}=n_{O_2 - consumed}\times M_{O_2}=0.1565\times32 = 5.008$ g.

Answer:

The mass of $SiO_2$ produced is approximately 18.81 g. The mass of the excess reactant ($O_2$) consumed is 5.008 g.