The given unbalanced equation is as follow,
Sn + H₃PO₄ → H₂ + Sn₃(PO₄)₄
In above equation first let get start with Sn, on left side its one and on right side its 3. So, multiply Sn on left side with 3 to balance Sn, Hence,
3 Sn + H₃PO₄ → H₂ + Sn₃(PO₄)₄
Now, balance Phosphorous, on left side it counts one and on right side it counts 4, so multiply H₃PO₄ on left side with 4 to balance P, Hence,
3 Sn + 4 H₃PO₄ → H₂ + Sn₃(PO₄)₄
Oxygen on both left and right are 16 in number, Hence it is balanced,
Hydrogen on left hand side are 12 in number, while that on right hand side only 2, So, multiply H₂ on right side by 6 to balance hydrogen atoms.
<u>3</u> Sn + <u>4</u> H₃PO₄ → <u>6</u><u> </u>H₂ + Sn₃(PO₄)₄
Above equation is balanced, and coefficients are highlighted bold and underlined.
Answer:
the water has greater salinity
Explanation:
The term "Salinity" refers to the amount of salt present in water. Recall that the amount of solute increases when the volume of solution decreases. This is because the amount of substance present in a solution depends on the volume of the solution.
Hence, as water evaporates, the volume of the solution decreases and the salinity (amount of salt present) increases.
Answer:
19.25g of C4H10
Explanation:
2C4H10
2×12×4+2×1×10
=116g
8CO2
=8×12+8×16×2
=352 g
116g of C4H10 produces 352 g of CO2
So, x g of C4H10 produces 58.4g of CO2
(cross multiply)
352 x=116×58.4
x=116×58.4÷352
x=19.25g of C4H10
Radiowave, microwave, infrared radiation
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
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