Question

reaction product 1/3

1. 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? input the numeric value here.
2. 2 co + o₂ → 2 co₂
3. 2 no + 2 co → n₂ + 2co₂
4. 2 geo + o₂ → 2 geo₂
5. 2 feo + o₂ → 2 feo₂
6. 2 sno + o₂ → 2 sno₂
7. 2 no + 2 geo → n₂ + 2geo₂
8. 2 sio + o₂ → 2 sio₂
9. 2 no + 2 sio → n₂ + 2 sio₂
10. 2 pbo + o₂ → 2 pbo₂
11. 2 nio + o₂ → 2 nio₂
12. 2 no + 2 sio → n₂ + 2 sio₂
13. 2 no + 2 sno → n₂ + 2 sno₂

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.
you must show work on the exams. practice showing the calculator answer and the sig figs answer.

Explanation:

Step1: Calculate moles of SiO

The molar - mass of SiO ($M_{SiO}$) is $M_{SiO}=28.09 + 16.00=44.09$ g/mol. The number of moles of SiO ($n_{SiO}$) is calculated using the formula $n=\frac{m}{M}$, where $m = 13.8$ g. So, $n_{SiO}=\frac{13.8\text{ g}}{44.09\text{ g/mol}}=0.313$ mol.

Step2: Calculate moles of $O_2$

The molar - mass of $O_2$ ($M_{O_2}$) is $M_{O_2}=2\times16.00 = 32.00$ g/mol. The number of moles of $O_2$ ($n_{O_2}$) is $n_{O_2}=\frac{13.8\text{ g}}{32.00\text{ g/mol}} = 0.431$ mol.

Step3: 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 0.431 mol of $O_2$, the amount of $SiO$ required is $n_{SiO - required}=2\times0.431$ mol = 0.862 mol. But we have only 0.313 mol of $SiO$. So, $SiO$ is the limiting reactant.

Step4: Calculate moles of $SiO_2$ produced

From the balanced equation, the mole ratio of $SiO$ to $SiO_2$ is 1:1. So, the number of moles of $SiO_2$ produced ($n_{SiO_2}$) is equal to the number of moles of $SiO$ reacted, which is 0.313 mol.

Step5: Calculate the mass of $SiO_2$ produced

The molar - mass of $SiO_2$ ($M_{SiO_2}$) is $M_{SiO_2}=28.09+2\times16.00=60.09$ g/mol. The mass of $SiO_2$ ($m_{SiO_2}$) is calculated using the formula $m = n\times M$. So, $m_{SiO_2}=0.313\text{ mol}\times60.09\text{ g/mol}=18.8$ g.

Answer:

18.8