calculate the volume of a 3.0 mol of oxygen, b 50.0g of hydrogen cloride gas at STP(molar volume of gas is 22.4dm³,molar mass of Hcl=36.5g)​

98L is the volume of the gas

Explanation:

To solve this question we need to find  the moles of O2 + moles HCl. With these moles and PV = nRT we can find the volume as follows:

Moles O2: 3.0mol

Moles HCl:

50.0g * (1mol/36.5g) = 1.37 moles HCl

Total moles:

3.0mol O2 + 1.37 mol HCl = 4.37 moles

PV = nRT

V = nRT/P

Where V is volume in L, n are moles = 4.37mol, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP, P = 1atm at STP

V = 4.37mol*0.082atmL/molK*273.15K / 1atm

V = 98L is the volume of the gas

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

c. is the answer

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