Answer: 12.5 moles NaNO3
Explanation: solution attached.
First balance predict the products
Na+ Mg( NO3)2 => Mg + NaNO3
Next balance the chemical equation
2 Na + Mg(NO3)2 => Mg + 2 NaNO3
Solve for the moles of Na
175 g Na x 1 mole Na/ 14 g Na
Next compare the mole ratios of Na and NaNO3 based from the balanced equation.
2 moles Na : 2 moles NaNO3
Ideal gas is defined as hypothetical gas that fits all the assumptions of the kinetic-molecular theory. There are several assumption statement that apply to the behavior of an ideal gas, First Gases consist of larger number of tiny particles that are far apart relative to their size, Second, collisions between gas particles and between particles and container walls are elastic collision, third, gas particles are in rapid, random motion and continuous, Fourth, the temperature of a gas depends on the average kinetic energy of the particles of the gas.
Answer:
![3.0molMg(OH)_2](https://tex.z-dn.net/?f=3.0molMg%28OH%29_2)
Explanation:
Hello there!
In this case, for these types of acid-base neutralizations, it is crucial to firstly set up the chemical reaction taking place between the acid and the base; in this case HCl and Mg(OH)2 respectively, whose products are obtained by switching around the anions and cations as shown below:
![HCl(aq)+Mg(OH)_2(aq)\rightarrow MgCl_2(aq)+H_2O(l)](https://tex.z-dn.net/?f=HCl%28aq%29%2BMg%28OH%29_2%28aq%29%5Crightarrow%20MgCl_2%28aq%29%2BH_2O%28l%29)
Which must be balanced to accurately predict the mole ratio on the reactants side:
![2HCl(aq)+Mg(OH)_2(aq)\rightarrow MgCl_2(aq)+2H_2O(l)](https://tex.z-dn.net/?f=2HCl%28aq%29%2BMg%28OH%29_2%28aq%29%5Crightarrow%20MgCl_2%28aq%29%2B2H_2O%28l%29)
Whereas we can see a 2:1 mole ratio of the acid to the base; thus, the moles of Mg(OH) required for the neutralization of 6.0 moles of HCl turn out to be:
![6.0molHCl*\frac{1molMg(OH)_2}{2molHCl} \\\\=3.0molMg(OH)_2](https://tex.z-dn.net/?f=6.0molHCl%2A%5Cfrac%7B1molMg%28OH%29_2%7D%7B2molHCl%7D%20%5C%5C%5C%5C%3D3.0molMg%28OH%29_2)
Best regards!
Answer:
the volume of the same gas at pressure of 1.00atm =737.3ml